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Physics 1101 Rotational Dynamics

    Moment of Inertia

  1. The moment of inertia of an oxygen molecule about an axis through the centre of mass and perpendicular to the line joining the atoms is 1.95 × 10-46 kg-m2. The mass of an oxygen atom is 2.66 × 10-26 kg. What is the distance between the atoms? Treat the atoms as particles.

    The moment of inertia for point particles is given by

    .

    Rewriting this for L yields

    .

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  2. An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness.
    (a) What is the mass of the lid/bottom?
    (b) What is the mass of the shell?
    (c) Find the moment of inertia of the can about the cylinder's axis of symmetry.

    (a) The mass has three components, M = Mlid + Mshell + Mbottom = 50 g, where the lid and the bottom are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom is that of a circle, Acircle = πr2 . The surface area of a cylindrical shell is Ashell = 2πrL. So the total area of can is Atotal = 2Acircle + Ashell = 2πrL + 2πr2 = 2πr(r + L). So the mass of the lid or bottom is given by . Thus Mlid = 5.392 g.

    (b) Similarly, the mass of the shell is given by . Thus Mshell = 39.216 g.

    (c) The total moment of inertia of the beer can is given by the sum of the individual pieces, Itotal = Ilid + Ishell + Ibottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have Itotal = 2×½Mlidr2 + Mshellr2 = 4.86 × 10-5 kg-m2.

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  3. A dumbbell consists of two uniform spheres of mass M and radius R joined by a thin rod of mass m, length L, and radius r (see diagram).
    (a) What is the moment of inertia about the centre of mass and perpendicular to the rod (Axis A)?
    (b) About an axis through one sphere and perpendicular to the rod (Axis B)?
    (c) About an axis along the rod (Axis C)?

    (a) The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, .For axis A, the rod is rotating about its centre of mass. Each sphere is a distance R+L/2 from the axis of rotation, so we must use the parallel axis theorem. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is . Thus we have

    (b) For axis B, the rod's centre is R+L/2 away from the axis of rotation. One sphere's centre is L+2R from the axis of rotation. The last sphere is rotating about axis B. Thus r


    (c) The bar and both spheres are rotating about their own centres when rotating about axis c. However, note that the bar is a cylinder or radius r in this configuration.



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  4. A hole of radius r has been drilled in a flat, circular plate of radius R. The centre of the hole is at a distance d from the centre of the circle. The mass of the complete body was M. Find the moment of inertia for rotation about an axis through the centre of the plate.

    We must treat the hole as an object of negative mass. The inertia of the object is then just. The plate and the hole are just disks and the inertia of a disk is . The hole is not rotating about its own centre of mass, so we must use the parallel axis theorem,

    .
    We are not yet finished; we do not know the mass of the hole. However since the plate is uniform, mass is proportional to area, so
    .
    Substituting this into our equation for the moment of inertia yields
    .

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  5. A thick spherical shell has an inner radius R1, an outer radius R2, and a mass M. The material that the spherical shell is made of is uniform. Find the moment of inertia of the thick shell about an axis through the centre of the sphere. The volume of a sphere is 4πr3/3.

    We must treat the hole at the centre as a sphere of negative mass. Since the moment of inertia of a sphere about its centre is , the moment of inertia of this object is

    .
    We are not yet finished since we do no know the mass of either the sphere or of the hole. We must find a way to relate these masses to the mass of the spherical shell, M. Fortunately the object is uniform so the mass of each is related to its volume. The volume of the spherical shell is simply the difference of its outer and inner volumes,
    .
    Thus the mass of the sphere is given by
    .
    Similarly, the mass of the hole is given by
    .
    Thus the moment of inertia of a thick spherical shell is
    .

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  6. The object shown in the diagram below consists of a 100-kg, 25.0 cm radius cylinder connected by four 5.00-kg, 0.75-m long thin rods to a thin-shelled outer cylinder of mass 20.0 kg. A small chunk of metal of mass 1.00 kg has been welded to the outer cylinder. What is the moment of inertia of the object about the centre of the inner cylinder? Treat the metal chunk as a point mass.

    The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The rods are not rotating about their centre of mass, so we must use the parallel axis theorem. The centre of each rod is Rcyl+L/2 from the axis of rotation at the centre of the object. The moments of inertia for a cylindrical shell, a disk, and a rod are MR2, , and respectively. The moment of inertia of a point mass is . Thus the total moment of inertia is:

    .

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  7. The object in the diagram below consists of five thin cylinders arranged in a circle. A thin disk has been welded to the tops and the bottoms of the cylinders. The cylinders each have a mass of 250 g, a length of 15.0 cm, and a radius of 1.00 cm. The disks are each 125 g and have a radius of 5.00 cm. Find the moment of inertia of the whole object about an axis through the centres of the disks.

    The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The cylinders are not rotating about their centre of mass, so we must use the parallel axis theorem. The centre of each rod is Rdisk- Rcyl from the axis of rotation at the centre of the object. The moments of inertia for a a disk or a cylindrical rod are is . The moment of inertia of a point mass is . Thus the total moment of inertia is:

    .

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    Pulleys

  8. Three point masses lying on a flat frictionless surface are connected by massless rods. Determine the angular acceleration of the body (a) about an axis through point mass A and out of the surface and (b) about an axis through point mass B. Express your answers in terms of F, L, and M. You will need to calculate the moment of inertia in each case.

    First we will calculate the moments of inertia. Since these are point masses we use the formula I = Σmi(ri)2:

    (a) IA = M(0)2 + 2M(L)2 + 3M(2L)2 = 14ML2;

    (b) Ib = M(L)2 + 2M(0)2 + 3M(L)2 = 4ML2.

    The angular acceleration is governed by the rotational form of Newton's Second Law, Στz = Izαz, where z is out of the paper in this problem and τz, Iz, and αz are all determined relative to the same axis.
    Axis A Axis B
    τz 2LF LF
    Iz 14ML2 4ML2
    Στz = Izαz 2LF = 14ML2αA LF = 4ML2αB

    So the acceleration about axis A is

    αA = F / 7ML ,

    and the acceleration about axis B is

    αB = F / 4ML .

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  9. The object in the diagram below is on a fixed frictionless axle. It has a moment of inertia of I = 50 kg-m2. The forces acting on the object are F1 = 100 N, F2 = 200 N, and F3 = 250 N acting at different radii R1 = 60 cm, R2 = 42 cm, and R3 = 28 cm. Find the angular acceleration of the object.

    Since the axle is fixed we only need to consider the torques and use Στz = Izαz. Each of the forces is tangential to the object, i.e R and F are at 90° to one another. Recall that clockwise torques are negative or into the paper in this case.

    Στz = Izαz
    -R1F1 + R2F2 + R3F3 = Iα

    So our equation for the acceleration is

    α = [-R1F1 + R2F2 + R3F3] / I .

    Substituting in the given values, α = 1.88 rad/s2.

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  10. A rope is wrapped around a solid cylindrical drum. The drum has a fixed frictionless axle. The mass of the drum is 125 kg and it has a radius of R = 50.0 cm. The other end of the rope is tied to a block, M = 10.0 kg. What is the angular acceleration of the drum? What is the linear acceleration of the block? What is the tension in the rope? Assume that the rope does not slip.

    Since the problem wants accelerations and forces, and one object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for each object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the drum has a fixed axle we need only consider the torques acting on it. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the block are weight and tension. Presumably the block will accelerate downwards. The only force directly acting on the drum which creates a torque is tension. Note that ropes, and therefore tensions, are always tangential to the object and thus normal to the radius. The other forces acting on the drum, the normal from the axle and the weight, both act through the CM and thus do not create torque. The drum accelerates counterclockwise as the block moves down.


    ΣFy = may ΣτΣτz = Izz
    T - Mg = -Ma RT = I

    Since the rope is wrapped around the drum, we also have the relationship a = Rα .

    Referring to the table of Moments of Inertia, we find that I = ½mR2 for a solid cylinder. So our first equation is T = Mg - Ma. Our second is RT = ½mR2(a/R), or when we simplify T = ½ma. Putting this result into the first equation yields a = Mg / [M + ½m] = 1.353 m/s2. Thus α = a/R = 2.706 rad/s2. As well, T = ½ma = ½mMg / [M + ½m] = 84.56 N.

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  11. Two blocks are connected over a pulley as shown below. The pulley has mass M and radius R. What is the acceleration of the blocks and the tension in the rope on either side of the pulley? (HINT: The tension must be different or the pulley would not rotate.)

    Since the problem wants accelerations and forces, and one object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for each object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the pulley has a fixed axle we need only consider the torques acting on it. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the block on the table are weight, a normal from the table, and tension. This block will accelerate to the right. The forces acting directing on the hanging block on the table are weight and tension. This block will accelerate downwards. The only forces directly acting on the pulley which creates torque are the tensions. Note that ropes, and therefore tensions, are always tangential to the object and thus normal to the radius. The tensions are different on either side of the pulley because of static friction - which we don't need to consider. The other forces acting on the pulley, the normal from the axle and the weight, both act through the CM and thus do not create torque. The pulley accelerates clockwise as the hanging block moves down.

    Table Block
    Pulley Hanging Block
    ΣFx = max ΣFy = may Στz = Izαz ΣFy = may
    T1 = m1a N - m1g = 0 RT1 - RT2 = -I T2 - m2g = -m2a

    From our table of Moments of Inertia, we find I = ½MR2 for a solid disk. As well, since the rope is strung over the pulley, we know a = Rα. Using these facts, the third equation becomes T1 - T2 = ½Ma. Using the first equation, T1 = m1a, and the second equation, T2 = m2g - m2a, we can eliminate T1 and T2 from the third equation:

    T1 - T2 = [m1a] - [m2g -m2a] = ½Ma .

    Collecting terms that contain a, and rearranging yields,

    a = m2g / [m1 + m2 + ½M].

    The angular acceleration of the pulley is thus

    α = a/R = m2g / [m1 + m2 + ½M]R .

    The tension in the left side of the rope is given by

    T1 = m1a = m1m2g / [m1 + m2 + ½M] .

    The tension in the hanging potion of the rope is

    T2 = m2g -m2a = m2g{1 - m2 / [m1 + m2 + ½M]}.

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  12. A winch has a moment of inertia of I = 10.0 kg-m2. Two masses M1 = 4.00 kg and M2 = 2.00 kg are attached to strings which are wrapped around different parts of the winch which have radii R1 = 40.0 cm and R2 = 25.0 cm.
    (a) How are the accelerations of the two masses and the pulley related?
    (b) Determine the angular acceleration of the masses. Recall that each object needs a separate free body diagram.
    (c) What are the tensions in the strings?

    Since the problem wants accelerations and forces, and one object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for each object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the pulley has a fixed axle we need only consider the torques acting on it. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the hanging blocks are weight and tension. Let's assume M1 accelerates downwards and thus M2 upwards. The only forces directly acting on the pulley which create torque are the tensions. Note that ropes, and therefore tensions, are always tangential to the object and thus normal to the radius. The tensions are different on either side of the pulley since they are different ropes. The other forces acting on the pulley, the normal from the axle and the weight, both act through the CM and thus do not create torque. The pulley accelerates counterclockwise as a result of our assumption for the acceleration of the blocks.

    Left Pulley Right
    ΣFy = may Στz = Izαz ΣFy = may
    T1 - M1g = -M1a1 R1T1 - R2T2 = Iα T2 - M2g = M2a2

    (a) The acceleration of M1 is equal to the tangential acceleration of the outside of the winch, so a1 = αR1. The acceleration of M2 is equal to the tangential acceleration of the inside ring of the winch, so a2 = αR2.

    (b) If we use the relationships from part (a), we can rewrite the equations in the table as

    T1 = M1g - M1R1α, and

    T2 = M2g + M2R2α.

    We use these results to eliminate T1 and T2 from the torque equation

    R1T1 - R2T2 = Iα

    R1[M1g - M1R1α] - R2[M2g + M2R2α] = Iα

    R1M1g - R2M2g = [I + M1(R1)2 + M2(R2)2] α.

    Thus we find

    α = g(R1M1 - R2M2)/[I + M1(R1)2 + M2(R2)2] = 1.002 rad/s2 .

    (c) Using this results, and our previous equations for the tension in each string, we find

    T1 = M1g - M1R1α = 38.60 N, and

    T2 = M2g + M2R2α = 20.12 N .

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  13. A rope connecting two blocks is strung over two real pulleys as shown in the diagram below. Determine the acceleration of the blocks and angular acceleration of the two pulleys. Block A is has mass of 10.0 kg. Block B has a mass of 6.00 kg. Pulley 1 is a solid disk, has a mass of 0.55 kg, and a radius of 0.12 m. Pulley 2 is a ring, has mass 0.28 kg, and a radius of 0.08 m. The rope does not slip.

    Since the problem wants accelerations and forces, and two objects rotate, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for each object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the pulleys have fixed axles, we need only consider the torques acting on each. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the hanging blocks are weight and tension. Let's assume A accelerates downwards and thus B upwards. The only forces directly acting on the pulleys which create torque are the tensions. Note that ropes, and therefore tensions, are always tangential to the object and thus normal to the radius. The tensions are different on either side of the pulley due to static friction with the surface of the pulleys. The other forces acting on each pulley, the normal from the axle and the weight, both act through the CM and thus do not create torque. The blocks are connected by the same rope and thus have the same magnitude of acceleration. The rope does not slip as it goes over the pulleys, so the pulleys have the same tangential acceleration as the rope. Since the pulleys have different radii, they have different angular accelerations.

    Left Disk Pulley Hoop Pulley Right
    ΣFy = may Στz = Izαz Στz = Izαz ΣFy = may
    T1 - MAg = -MAa RT1 - RT2 = Idiskα1 R2T2 - R2T3 = Ihoopα2 T3 - MBg = MBa

    In addition to the equations we have found above, we also know that the tangential acceleration of the pulleys is the same as the acceleration of the rope. Thus the angular acceleration of each pulley is related to a by α1 = a/R1 and α2 = a/R2. Examining a table of Moments of Inertia reveals that Idisk = ½Mdisk(R1)2 and Idisk = Mhoop(R2)2. Using this information allows us to rewrite the equations as

    T1 = MAg - MAa      (1),

    T1 - T2 = ½Mdiska     (2),

          T2 - T3 = Mhoopa      (3), and

    T3 =MBg + MBa       (4).

    If we add equations (2) and (3) together, we get

    T1 - T3 = (½Mdisk + Mhoop)a .

    Then equations (1) and (4) can be used to eliminate T1 and T3 from the above

    [MAg - Maa] - [MBg + Mba] = (½Mdisk + Mhoop)a .

    Collecting terms involving a and rearranging yields,

    a = (MA - MB)g / (MA + MB + ½Mdisk + Mhoop) = 2.370 m/s2.

    Using the above result, we find the angular accelerations

    α1 = a/R1 = 19.75 rad/s2

    and α2 = a/R2 = 29.63 rad/s2 .

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    Rolling Objects

  14. A yo-yo has a mass M, a moment of inertia I, and an inner radius r. A string is wrapped around the inner cylinder of the yo-yo. A person ties the string to his finger and releases the yo-yo. As the yo-yo falls, it does not slip on the string (i.e. the yo-yo rolls). Find the acceleration of the yo-yo.

    Since the problem wants an acceleration, and an object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for each object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the yo-yo does not have a fixed axle, we need consider the torques acting about the CM. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the hanging yo-yo are weight and tension. Let's assume it accelerates downwards. The only force directly acting on the pulley which creates a torque is the tensions, the weight acts from the CM and cannot create a torque. Note that strings, and therefore tensions, are always tangential to the object and thus normal to the radius.

    The yo-yo is said to roll without slipping. That phrase means that the angular acceleration of the yo-yo about its CM is related to its linear acceleration by a = Rα. Note that since the string is tied around the inner cylinder; it is that radius which figures into the relation.

    ΣFy = may Στz = Izαz
    T - mg = -ma rT = Iαcm

    Since αcm = a/r, we have two simple equations

    T = mg - ma, and T = Ia / r2.

    Eliminating T from the first equation yields,

    a = mg/[m+I/r2] .

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  15. A solid cylinder rolls down an inclined plane without slipping. The incline makes an angle of 25.0 to the horizontal, the coefficient of static friction is μs = 0.40, and Icyl = ½MR2. Hint - you may not assume that static friction is at its maximum!
    (a) Find its acceleration.
    (b) Find the angle at which static friction is at its maximum, at just above this angle the object will start to slip.

    Since the problem wants an acceleration, and an object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for the object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the ball does not have a fixed axle, we need consider the torques acting about the CM. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directly on the ball are the normal, weight, and friction. Naturally the cylinder will accelerate downward the incline. Since the cylinder isn't slipping, its forward rate of rotation, its angular acceleration, is also forward. The only force directly acting on the cylinder which creates a torque is the friction, the normal and the weight act through the CM and cannot create a torque. Note that friction, being along the surface, is tangential to the cylinder and thus normal to the radius.

    The cylinder is said to roll without slipping. That phrase means that the angular acceleration of the ball about its CM is related to its linear acceleration by a = Rα.

    The type of friction is static since we are told that the cylinder is rolling without slipping. The only point left to resolve is in which direction it points. Since friction creates the only torque, and we have decided that α is forward, then friction must be up the incline. Only point to be careful about is that in rolling problems, one seldom is dealing with the fs MAX unless it is explicitly stated.

    ΣFx = max ΣFy = may Στz = Izαz
    mgsinθ - fs = ma N - mgcosθ = 0 -Rfs = -Icylαcm

    (a) The first equation is mgsinθ - ma = fs. The third equation can be simplified by using the given value for Icyl and by noting that αcm = a/R. Then the third equation becomes

    fs = ½ma .

    This can be substituted into the first equation to get

    mgsinθ - ma = ½ma .

    Solving for a yields

    a = (2/3)gsinθ = 2.764 m/s2.

    (b) We can use this result with fs = ½ma, to get an expression for fs,

    fs = (1/3)mgsinθ .      (1)

    However, the maximum value of fs is μsN. The second equation gives N = mgcosθ, so

    fs = μsmgcosθ .       (2)

    Using (1) and (2) to eliminate fs, yields

    (1/3)mgsinθ = μsmgcosθ .

    Using the identity tanθ = sinθ/cosθ, we get

    θ = tan-1(3) = 50.2° .

    This is the angle at which the cylinder would start to slip as it moved down the incline.

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  16. A thin-shelled cylinder rolls up an inclined plane without slipping. The incline makes an angle of 25.0 to the horizontal, the coefficient of static friction is μs = 0.40, and Ihoop = MR2.
    (a) Find its acceleration.
    (b) Find the angle which the object will start to slip.

    Since the problem wants an acceleration, and an object rotates, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for the object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the hoop does not have a fixed axle, we need to consider the torques acting about the CM. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directing on the hoop are the normal, weight, and friction. As the hoop goes up the incline it is slowing down, so its acceleration is down the incline. Since the hoop isn't slipping, its forward rate of rotation, its angular acceleration, is decreasing. The only force directly acting on the hoop which creates a torque is the friction, the normal and the weight act through the CM and cannot create a torque. Note that friction, being along the surface, is tangential to the hoop and thus normal to the radius.

    The hoop is said to roll without slipping. That phrase means that the angular acceleration of the hoop about its CM is related to its linear acceleration by a = Rα.

    The type of friction is static since we are told that the hoop is rolling without slipping. The only point left to resolve is in which direction it points. Since friction creates the only torque, and we have decided that α is backward, then friction must be up the incline. One point to be careful about is that in rolling problems, one seldom is dealing with the fs MAX unless it is explicitly stated.

    ΣFx = max ΣFy = may Στz = Izαz
    mgsinθ - fs = ma N - mgcosθ = 0 -Rfs = -Ihoopαcm

    (a) The first equation is mgsinθ - ma = fs. The third equation can be simplified by using the given value for Ihoop and by noting that αcm = a/R. Then the third equation becomes

    fs = ma .

    This can be substituted into the first equation to get

    mgsinθ - ma = ma .

    Solving for a yields

    a = ½gsinθ = 4.146 m/s2 .

    (b) We can use this result with fs = ma, to get an expression for fs,

    fs = ½mgsinθ .      (1)

    However, the maximum value of fs is μsN. The second equation gives N = mgcosθ, so

    fs = μsmgcosθ .      (2)

    Using (1) and (2) to eliminate fs, yields

    ½mgsinθ = μsmgcosθ .

    Using the identity tanθ = sinθ/cosθ, we get

    θ = tan-1(2μ) = 38.7° .

    This is the angle at which the hoop would start to slip as it moved down the incline.

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  17. A toy car has a frame of mass M and four wheels of mass m. The wheels are solid disks. The car is placed on an incline and let go. Assume each tire supports one-quarter of the car's weight.
    (a) Find the acceleration of the toy car.
    (b) If the coefficient of static friction is μ, find an expression for the angle at which the wheels begin to slip.

    Since the problem wants an acceleration, and objects rotate, that suggests we must use both the linear and rotational versions of Newton's Second Law. Applying Newton's Second Law requires that we draw free body diagrams for the object. In particular for any rotating body we must draw an extended FBD in order to calculate the torques. Since the wheel axle is not fixed, we need consider the torques acting about each wheel's CM. Once the diagrams are drawn, we use ΣFx = max, ΣFy = may, and Στz = Izαz to get a set of equations.

    The forces acting directly on each wheel are the normal, weight, and friction. Naturally the car will accelerate downward the incline. We will assume that each wheel supports one quarter of the car's weight and thus are all identical - we need only one FBD. Since the wheel isn't slipping, its forward rate of rotation, its angular acceleration, is also forward. The only force directly acting on the wheel which creates a torque is the friction, the normal and the weight act through the CM and cannot create a torque. Note that friction, being along the surface, is tangential to the wheel and thus normal to the radius.

    The wheel is said to roll without slipping. That phrase means that the angular acceleration of the wheel about its CM is related to its linear acceleration by a = Rα.

    The type of friction is static since we are told that the wheel is rolling without slipping. The only point left to resolve is in which direction it points. Since friction creates the only torque, and we have decided that α is forward, then friction must be up the incline. One point to be careful about is that in rolling problems, one seldom is dealing with the fs MAX unless it is explicitly stated.

    ΣFx = max ΣFy = may Στz = Izαz
    (m+¼M)gsinθ - fs = (m+¼M)a N - (m+¼M)gcosθ = 0 Rfs = Idiskαcm

    (a) The first equation is mgsinθ - ma = fs. The third equation can be simplified by using Ihoop = ½mR2 and by noting that αcm = a/R. Then the third equation becomes

    fs = ½ma .

    This can be substituted into the first equation to get

    (m+¼M)gsinθ - ½ma = (m+¼M)a .

    Solving for a yields

    a = gsinθ[(M+4m)/(M+6m)] .

    (b) We can use this result with fs = ½ma, to get an expression for fs,

    fs = ½mgsinθ[(M+4m)/(M+6m)] .     (1)

    However, the maximum value of fs is μsN. The second equation gives N = (m+¼M)gcosθ, so

    fs = μs(m+¼M)gcosθ .             (2)

    Using (1) and (2) to eliminate fs, yields

    ½mgsinθ[(M+4m)/(M+6m)] = μs(m+¼M)gcosθ .

    Using the identity tanθ = sinθ/cosθ, we get

    θ = tan-1([M+6m]/2m) .

    This is the angle at which the wheels would start to slip as it moved down the incline.

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  18. A person pulls a heavy lawn roller by the handle with force F so that it rolls without slipping. The handle is attached to the axle of the solid cylindrical roller. The handle makes an angle θ to the horizontal. The roller has a mass of M and a radius R. The coefficients of friction between the roller and the ground are μs and μk.
    (a) Find the acceleration of the roller.
    (b) Find the frictional force acting on the roller.
    (c) If the person pulls too hard, the roller will slip. Find the value of F at which this occurs.

    First we draw the FBD. Clearly we have F, weight, and a normal force acting on the roller. As well, there must be static friction since we have rolling without slipping. It is not at a maximum since the roller only starts to slip in part (c). The direction of fs must be to the right, since the force F pulls the roller into the ground, the ground pushes back. We assume the accelerations are as shown, left and ccw. Note that these are consistent.

    We apply Newton's Second Law to the problem.

    Σ Fx

    = max

    Σ Fy

    = may

    Σ τcm

    = Icmα cm

    − Fcos(θ ) + fs

    = − Ma

    N + Fsin(θ ) − Mg

    = 0

    Rfs

    = Iα

    We also know that a = Rα and I = ½MR2. We substitute these relationships into the torque equation

    Rfs = ½MR2(a/R)

    (1)

    This yields an equation for fs,

    fs = ½Ma

    (2)

    We take this result and substitute it into the x-component equation

    − Fcos(θ ) + ½Ma = − Ma

    (3)

    Solving for a, as required in part (a), yields

    a = (2/3)(F/M) cos(θ )

    (4)

    We can also find fs, as required in part (b), by substituting (4) back into Eqn. (2);

    fs = (1/3)Fcos(θ )

    (5)

    Part (c) asks us when the roller will slip. This occurs when fs = fsmax. Now we know fsmax = μsN. We find N from the y-component equation, so

    fsmax = μs[Mg − Fsin(θ )]

    (6)

    Equating Eqns. (5) and (6) will tell us the value of F at which slipping occurs

    (1/3)Fcos(θ )= μs[Mg − Fsin(θ )]

    (7)

    We get F by itself on the left-hand side

    F[(1/3)cos(θ ) + μssin(θ )] = μsMg

    (8)

    So

    (9)

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  19. A yo-yo of Mass M, moment of inertia I, and inner and outer radii r and R, is gently pulled by a string with tension T as shown in the diagram below. The coefficients of friction between the yo-yo and the table are μs and μk.
    (a) Find the acceleration.
    (b) Find the friction acting on the yo-yo.
    (c) At what value of T will the yo-yo begin to slip?

    We know there is a normal and weight acting on the yo-yo but these do not create torques as they operate on or through the Centre of Mass. There is non-maximum static friction acting but we must determine it's direction. First if there were no T, there would be no friction. T is twisting the yo-yo counterclockwise pushing the yo-yo into the surface. The surface reacts by pushing back. The FBD looks like

    Here I have guessed that the yo-yo will roll backwards. Notice that my choice of a and α are consistent.

    I apply Newton's Second Law

    Σ Fx

    = max

    Σ Fy

    = may

    Σ τcm

    = Icmacm

    T − fs

    = − Ma

    N − Mg

    = 0

    rT − Rfs

    = Iα

    We also know a = Rα .

    The torque equation becomes can be solved for fs

    fs = (r/R)T − (I/R2)a .

    (1)

    We substitute this into the x-component equation

    T − [(r/R)T − (I/R2)a] = − Ma

    (2)

    We bring the term involving a from the left to the right and solve for a in terms of T,

    T(R-r)/R = − (M + I/R2)a

    (3)

    or

    (4)

    The fact that a is negative tells me that my guess about the direction of a and α are wrong. The acceleration is forward and counterclockwise.

    Substituting Eqn. (4) into Eqn. (1), we find

    (5)

    Since fs is positive, it must be have been chosen in the right direction.

    Now as we see from Eqn. (5), as T increases so does fs. We know fs £ μsN or fs £ μsMg in our case when we make use of the y-component equation. Thus we have a limit on T

    (6)

    This yields the result

    (7)

    If T is any bigger, the yo-yo will slip.

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  20. Rotational Work and Energy

  21. A rope is wrapped abound a cylindrical drum as shown below. It is pulled with a constant tension of 100 N for six revolutions of the drum. The drum has a radius of 0.500 m. A brake is also applying a force to the drum. The brake pushes inwards on the drum with a force of 200 N. The pressure point is 0.350 m from the centre of the drum. The coefficient of kinetic friction between the brake and the drum is 0.50. Determine the work done by each torque.

    Work is defined by the formula W = τΔφ in rotational cases. Since the rope does not slip as it is pulled, the object rotates 6 times clockwise or Δφ = -12π. We know fk = μkN. In this problem, N equals how hard the brake is pressed. Note that the tension and the friction are tangential to the drum.

    (a) WT = (-RT)Δφ = -(0.5m)(100 N)(-12π) = 1.88 × 103 J.

    (b) Wf = (rfk)Δφ = (0.35 m)(0.50 × 200 N)(-12π) = -1.32 × 103 J.

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  23. A rope is wrapped exactly three times around a cylinder with a fixed axis of rotation at its centre. The cylinder has a mass of 250 kg and a diameter of 34.0 cm. The rope is pulled with a constant tension of 12.6 N. The moment of inertia of a cylinder about its centre is I = ½MR2. (a) What is the work down by the rope as it is pulled off the cylinder. Note that the rope does not slip. (b) If the cylinder was initially at rest, what is its final angular velocity? Note that ropes are always tangential to the surfaces that they are wrapped around. Note that the work done by the tension is non-conservative.

    Since the problem involves a change is speed, we make use of the Generalized Work-Energy Theorem. Since there is only a change in rotational speed,

    WNC = ΔE = Kf - Ki = ½I[(ωf)2 - (ω0)2] = ½I(ωf)2 .

    The nonconservative force in this problem is tension, WNC = WT. So we have

    WT = ½I(ωf)2 .           (1)

    (a) The definition of work in rotational situations is W = τΔφ. Tension is always tangential to cylinders so τT = -RT. Since the rope does not slip, the cylinder rotates clockwise three times so Δφ = -6π. We can thus find the work done by the tension

    WT = (-RT)(-6π) = (0.34 m)(12.6 N)6π = 40.4 J .

    (b) Then we find the final velocity from equation (1),

    ωf = [2WT/I]½ .

    According to the table of Moments of Inertia, I = ½MR2 for a solid cylinder. So

    ωf = [4WT/MR2]½ = 6.10 rad/s .

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  24. A large cylinder of mass M = 150 kg and radius R = 0.350 m. The axle on which the cylinder rotates is NOT frictionless. A rope is wrapped around the cylinder exactly ten times. From rest, the rope is pulled with a constant tension of 25.0 N. The rope does not slip and when the rope comes free, the cylinder has a forward angular velocity f = 10.5 rad/s. The moment of inertia of a cylinder is I = ½MR2.
    (a) What angle was the cylinder rotated through?
    (b) What is the frictional torque of the axle?
    (c) How long will it take the frictional torque to bring the cylinder to a stop?
    (d) How many revolutions will it have turned?

    (a) Since the rope does not slip, the cylinder rotates ten times so Δφ = -20π, where the rotation is assumed to be clockwise.

    (b) Since the problem involves forces and a change is rotational speed, we make use of the Generalized Work-Energy Theorem. Since there is only a change in rotational kinetic energy,

    WNC = ΔE = Kf - Ki = ½I[(ωf)2 - (ω0)2] = ½I(ωf)2 .

    The nonconservative forces in this problem are the tension and the axle friction, WNC = WT + Wf. So we have

    WT + Wf = ½I (ωf)2 .           (1)

    The definition of work in rotational situations is W = τΔφ. Tension is always tangential to cylinders so τT = -RT, again assuming the rope pulls the cylinder clockwise. Thus the work done by the tension is

    WT = (-RT)(-20π) = (0.35 m)(25 N)(20) = 549.78 J .

    Using this result and equation (1), we can find the work done by friction

    Wf = ½I(ωf)2 - WT = ¼M(Rf)2 - WT = 506.46 J - 549.78 J = -43.3178 J .

    Since Wf = τf Δφ, we find the frictional torque to be

    τf = Wf / Δφ = Wf / -20π = +0.6894 Nm ;

    the sign indicating that it is counterclockwise.

    (c) After the rope comes off the cylinder, the only force acting is friction so the generalized Work-Energy Theorem becomes

    Wf = ΔE = Kf - Ki = ½I[(ωf) 2 - (ω0)2] = -½I(ω0)2 ,

    where ω0 here is ωf from the first part of the problem and the new ωf = 0 since the drum comes to a stop. Again Wf = τfΔφ, where τf is the result from part (b). Therefore

    τfΔφ = -½I(ω0)2 = -(506.46 J) / 0.6894 Nm = -734.6 rad = 117 rev clockwise.

    (d) To find the time it takes to slow down, note that we have the initial and final angular velocities and the angular displacement. Referring to our kinematics equations, we find

    Δφ = ½(ωf + ω0)t.

    Rearranging for t yields,

    t = 2Δφ / (ωf + ω0) = 140 s .

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  25. A solid ball, a cylinder, and a hollow ball all have the same mass m and radius R. They are allowed to roll down a hill of height H without slipping. How fast will each be moving on the level ground?
  26. The only external force on the object, excluding gravity which is taken into account through gravitational potential energy, is static friction. For objects which roll across a stationary surface, static friction does no work, so here 0 = ΔE, where E is the observable or mechanical energy.

    As the object drops down the hill, it loses gravitational potential energy while it gains both linear and rotational kinetic energy. Thus we have

    0 = -mgH + ½mv2 + ½Iω2

    Now since the object rolls without slipping, the angular velocity ω is related to the linear velocity v by ω = v/R. So our equation becomes

    0 = -mgH + ½mv2 + ½I(v/R)2

    Isolating v2 terms, we find v2[m + I/R2] = 2mgH. Solving for v we get 

    To proceed further, we need to know the moment of inertia of each object about its CM. We consult a table of moments of inertia and find

    Shape ICM v
    solid sphere 2/5MR2
    solid cylinder ½MR2
    hollow sphere 2/3MR2

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  27. A cylinder of mass M and radius R, on an incline of angle θ, is attached to a spring of constant K. The spring is not stretched. Find the speed of the cylinder when it has rolled a distance L down the incline.

  28. The only external force on the system, excluding gravity which is taken into account through gravitational potential energy, is static friction. For objects which roll across a stationary surface, static friction does no work, so here 0 = ΔE, where E is the observable or mechanical energy.

    As the object rolls distance L down the incline, it loses gravitational potential energy while it gains both linear and rotational kinetic energy. Moreover, the spring gains Spring Potential Energy. Thus we have

    0 = -mgH + ½mv2 + ½Iω2 + ½KL2

    The relationship between H and the L is H = Lsinθ. Since the object rolls without slipping, the angular velocity ω is related to the linear velocity v by ω = v/R. Furthermore, consulting a table of values, the moment of inertia of a cylinder about the perpendicular axis through the CM, is ½MR2. Our equation becomes

    0 = -mgLsinθ + ½mv2 + ½(½MR2)(v/R)2 + ½KL2

    Isolating v2 terms, we find v2[½m + ¼m] = mgLsinθ - ½KL2. Solving for v we get

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  29. A block of mass m is connected by a string of negligible mass to a spring with spring constant K which is in turn fixed to a wall. The spring is horizontal and the string is hung over a pulley such that the mass hangs vertically. The pulley is a solid disk of mass M and radius R. As shown in the diagram below, the spring is initially in its equilibrium position and the system is not moving.
    (a) Use energy methods, to determine the speed v of the block after it has fallen a distance h. Express your answer in terms of g, m, K, M, and h.
    (b) The block will oscillate between its initial height and its lowest point. At its lowest point, it turns around. Use your answer to part (a) to find where it turns around.
    (c) Use your answer to part (a) and calculus to find the height at which the speed is a maximum.
  30. (a) The problem involves a change in height and speed and has a spring, so we would apply the generalized Work-Energy Theorem even if not directed to do,

    WNC = ΔE = (Kf - Ki) + (Uf - Ui) ,            (1)

    where K is the sum of all the linear and rotational kinetic energies of each object, and U is the sum of the spring and gravitational potential energies. Since there is no kinetic friction acting on the system, WNC = 0.

    Examining the problem object by object we see that the spring stretches, so there is an increase in spring potential energy. The pulley starts rotating, so there is an increase in its rotational kinetic energy. The block drops, so there is a decrease in its gravitational potential energy. As well, as the block drop, it increases its kinetic energy. Equation (1) for this problem is thus

    0 = ½Kx2 + ½Iω 2 - mgh + ½mv2 .

    Since the spring is connected to the block, the spring stretches as much as the block drops, so x = h. We are told that the pulley is a solid disk, so I = ½MR2. Since the rope does not slip the tangential speed of the pulley is the same as the rope and thus ω = v/R. Substituting these relations back into our equation yields,

    0 = ½Kh2 + ¼Mv2 - mgh + ½mv2 .

    Collecting the terms with v, and solving for v yields

    v = [(2mgh - kh2) / (m + M/2)]½.            (2)

    (b) Recall from our discussions on kinematics that an object turns around when its velocity is zero. Setting equation (2) to zero

    [(2mgh - kh2) / (m + M/2)]½ = 0 ,

    we see that the numerator is zero when

    2mgh - kh2 = 0 .

    Solving this for h reveals that the object turns around when h = 2mg/k.

    (c) To find the maximum velocity, we need to find dv/dh = 0. Taking the derivative of equation (2) yields

    dv/dh = ½{1/[(2mgh - kh2) / (m + M/2)] ½}[2mg-2kh]/[m+M/2] = 0 .

    We see that the numerator to zero when

    2mg-2kh = 0 .

    Solving this for h reveals that the object turns around when h = mg/k, halfway between the starting position and the turnaround point.

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  31. A system of weights and pulleys is assembled as shown below. The pulleys are all fixed. The pulleys on the sides are disks of mass md and radius Rd; the central pulley is a hoop of mass mh and radius Rh. A massless ideal rope passes around the pulleys and joins two weights. The rope does not slip. The two weights have masses m1 and m2 respectively. Use energy methods to find the velocity of the weights as a function of displacement, x. The moment of inertia of a disk is Idisk = ½MR2 and of a hoop is Ihoop = MR2.

    The problem involves changes in height, speed, and rotation, so we would apply the generalized Work-Energy Theorem even if not directed to do,

    WNC = ΔE = (Kf - Ki) + (Uf - Ui) ,           (1)

    where K is the sum of all the linear and rotational kinetic energies of each object, and U is the sum of the gravitational potential energies. Since there is no kinetic friction acting on the system, WNC = 0.

    Let's assume block 2 moves down. Thus block 1 moves up an identical amount h. Since block 2 drops, there is a decrease in its gravitational potential energy. Block 1 will increase its potential energy. As well, as both blocks move, they increases its kinetic energy. The pulleys start rotating, so there is an increase in the rotational kinetic energy of each. Equation (1) for this problem is thus

    0 = -m2gh + m1gh + ½m2v2 + ½m1v2 + 2×½IDD)2 + ½IHH) 2 .

    We are told that Idisk = ½MD(RD)2 and Ihoop = MH(RH)2. Since the rope does not slip the tangential speed of the pulleys is the same as the rope and thus ω = v/R. Substituting this relations back into our equation yields,

    0 = -(m2 - m1)gh + ½m2v2 + ½m1v2 + ½MD(RD)2(v/RD)2 + ½ MH(RH)2(v/RH)2 .

    Collecting the terms with v, and solving for v yields

    v = [2(m2 - m1)gh / (m1 + m2 + MD + MH)]½.

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  32. In the diagram block M1 is connected to M2 by a very light string running over three identical pulleys. The pulleys are disks with mass Mp and the rope does not slip. The coefficient of kinetic friction for the horizontal surface that M1 is on is μk. Find an expression for the speed v of block M2 in terms of the distance L that block M1 moves to the right. Your answer should be expressed in terms of L, M1, M2, Mp, μk, and g only.

  33. The problem involves changes in height, speed, and rotation, so we would apply the generalized Work-Energy Theorem even if not directed to do so,

    WNC = ΔE = (Kf - Ki) + (Uf - Ui) ,           (1)

    where K is the sum of all the linear and rotational kinetic energies of each object, and U is the sum of the gravitational potential energies. Since there is kinetic friction acting on the system, WNC = Wf. To find Wf, we need to the frictional force. To find a force, we draw a FBD and use Newton's Second Law.

    i
    j
    ΣFx = max ΣFy = may
    T - fk = M1a N - M1g = 0

    The second equation gives N = M1g and we know fk = μk N, so fk = μkM1g. Therefore, the work done by friction is Wf = -fk Δx = -μ kM1gΔx.

    Next consider the change in energy of each object. M1 increases its linear kinetic energy. M2 also increases its linear kinetic energy but loses gravitational potential energy. The three identical pulleys increase their rotational kinetic energies. Thus equation (1) becomes

    kM1gΔx = ½M1v2 + ½M2v2 - M2gh + 3×½Idisk ω2 .

    Since M1 and M2 are connected by the same rope, they have the same speed and move the same distance so that Δx = h. Consulting a table of Moments of Inertia, we find Idisk = ½MPR2. Since the rope does not slip, the tangential speed of the pulleys is the same as the rope and thus ω = v/R. Substituting this relations back into our equation yields,

    kM1gh = ½M1v2 + ½M2v2 - M2gh + 3×½[½MPR2] (v/R)2 .

    Collecting the terms with v yields

    ½[M1 + M2 + (3/2)MP] v2 = [M2 - μ kM1]gh .

    Solving for v yields,

    v = [2(M2 - μ kM1)gh / (M1 + M2 + (3/2)MP)]½ .

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  34. A block of mass M on a flat table is connected by a string of negligible mass to a vertical spring with spring constant K which is fixed to the floor. The string goes over a pulley that is a solid disk of mass M and radius R. As shown in the diagram below, the spring is initially in its equilibrium position and the system is not moving. A person pulls the block with force F through a distance L. Determine the speed v of the block after it has moved distance L.The tabletop is frictionless.

    The problem involves changes in height, speed, and rotation, so we would apply the generalized Work-Energy Theorem even if not directed to do so, 

    Wext = ΔE, (1)

    where E is the sum of all the mechanical energies of each object. If the system consists of the spring, string, pulley, block and the earth, then F is an external force acting on the system and Wext = FL.

    Next consider the change in energy of each object. The spring stretches as so increases its potential energy. The pulley turns from rest so increases its rotational kinetic energy. The block moves from rest so it increases its linear kinetic energy. Thus equation (1) becomes

    FL = ½Kx2 + ½Idiskω2 + ½Mv2

    Since the block and spring are connected by the same string, the spring has stretched x = L. Consulting a table of Moments of Inertia, we find Idisk = ½MR2. Since the string does not slip, the tangential speed of the pulleys is the same as the rope and thus ω = v/R. Substituting this relations back

    FL = ½KL2 + ½[½MR2](v/R)2 + ½Mv2 .

    Collecting the terms with v yields

    3/4Mv2 = FL – ½KL2 . 

    Solving for v yields,

    v = [(4M – 2KL2) / 3M]½ .

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