Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
Physics 1101  Rotational Dynamics 
The moment of inertia for point particles is given
by
Rewriting this for L yields
(a) The mass has three components, M = M_{lid} + M_{shell} + M_{bottom} = 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, A_{circle} = πr^{2} . The surface area of a cylindrical shell is A_{shell} = 2πrL. So the total area of can is A_{total} = 2A_{circle} + A_{shell} = 2πrL + 2πr^{2} = 2πr(r + L). So the mass of the lid or bottom is given by . Thus M_{lid} = 5.392 g.
(b) Similarly, the mass of the shell is given by . Thus M_{shell} = 39.216 g.
(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, I_{total} = I_{lid} + I_{shell} + I_{bottom}. 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 I_{total} = 2×½M_{lid}r^{2} + M_{shell}r^{2} = 4.86 × 10^{5} kgm^{2}.
(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.
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 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
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 R_{cyl}+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 MR^{2}, , and respectively. The moment of inertia of a point mass is . Thus the total moment of inertia is:
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 R_{disk} R_{cyl} 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:
First we will calculate the moments of inertia. Since these are point masses we use the formula I = Σm_{i}(r_{i})^{2}:
(a) I_{A} = M(0)^{2} + 2M(L)^{2} + 3M(2L)^{2 }= 14ML^{2};
(b) I_{b} = M(L)^{2} + 2M(0)^{2} + 3M(L)^{2 }= 4ML^{2}.
The angular acceleration is governed by the rotational form of Newton's Second Law, Στ_{z} = I_{z}α_{z}, where z is out of the paper in this problem and τ_{z}, I_{z}, and α_{z} are all determined relative to the same axis.
Axis A  Axis B  
τ_{z}  2LF  LF 
I_{z}  14ML^{2}  4ML^{2} 
Στ_{z} = I_{z}α_{z}  2LF = 14ML^{2}α_{A}  LF = 4ML^{2}α_{B} 
So the acceleration about axis A is
and the acceleration about axis B is
Since the axle is fixed we only need to consider the torques and
use Στ_{z}
= I_{z}α_{z}. 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.
R_{1}F_{1} + R_{2}F_{2} + R_{3}F_{3} = Iα 
So our equation for the acceleration is
Substituting in the given values, α = 1.88 rad/s^{2}.
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z} 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.
ΣF_{y} = ma_{y}  ΣτΣτ_{z} = I_{zz} 
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 = ½mR^{2} for a solid cylinder. So our first equation is T = Mg  Ma. Our second is RT = ½mR^{2}(a/R), or when we simplify T = ½ma. Putting this result into the first equation yields a = Mg / [M + ½m] = 1.353 m/s^{2}. Thus α = a/R = 2.706 rad/s^{2}. As well, T = ½ma = ½mMg / [M + ½m] = 84.56 N.
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 ΣF_{x}
= ma_{x}, ΣF_{y} =
ma_{y}, and Στ_{z}
= I_{z}α_{z} 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.
Pulley  Hanging Block  
ΣF_{x} = ma_{x}  ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z}  ΣF_{y} = ma_{y} 
T_{1} = m_{1}a  N  m_{1}g = 0  RT_{1}  RT_{2} = I  T_{2}  m_{2}g = m_{2}a 
From our table of Moments of Inertia, we find I = ½MR^{2} 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 T_{1 } T_{2} = ½Ma. Using the first equation, T_{1} = m_{1}a, and the second equation, T_{2} = m_{2}g  m_{2}a, we can eliminate T_{1} and T_{2} from the third equation:
Collecting terms that contain a, and rearranging yields,
The angular acceleration of the pulley is thus
The tension in the left side of the rope is given by
The tension in the hanging potion of the rope is
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z} to get a
set of equations.
The forces acting directing on the hanging blocks are weight and tension. Let's assume M_{1} accelerates downwards and thus M_{2} 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 
ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z}  ΣF_{y} = ma_{y} 
T_{1}  M_{1}g = M_{1}a_{1}  R_{1}T_{1}  R_{2}T_{2} = Iα  T_{2}  M_{2}g = M_{2}a_{2} 
(a) The acceleration of M_{1} is equal to the tangential acceleration of the outside of the winch, so a_{1} = αR_{1}. The acceleration of M_{2} is equal to the tangential acceleration of the inside ring of the winch, so a_{2} = αR_{2}.
(b) If we use the relationships from part (a), we can rewrite the equations in the table as
We use these results to eliminate T_{1} and T_{2}
from the torque equation
Thus we find
(c) Using this results, and our previous equations for the tension in each string, we find
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z}
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 
ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z}  Στ_{z} = I_{z}α_{z}  ΣF_{y} = ma_{y} 
T_{1}  M_{A}g = M_{A}a  RT_{1}  RT_{2} = I_{disk}α_{1}  R_{2}T_{2}  R_{2}T_{3} = I_{hoop}α_{2}  T_{3}  M_{B}g = M_{B}a 
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 α_{1} = a/R_{1} and α_{2} = a/R_{2}. Examining a table of Moments of Inertia reveals that I_{disk} = ½M_{disk}(R_{1})^{2} and I_{disk} = M_{hoop}(R_{2})^{2}. Using this information allows us to rewrite the equations as
T_{1}  T_{2} = ½M_{disk}a (2),
T_{2}  T_{3} = M_{hoop}a (3), and
T_{3} =M_{B}g + M_{B}a (4).
If we add equations (2) and (3) together, we get
Then equations (1) and (4) can be used to eliminate T_{1}
and T_{3} from the above
Collecting terms involving a and rearranging yields,
Using the above result, we find the angular accelerations
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 yoyo does not have a fixed
axle, we need consider the torques acting about the CM. Once
the diagrams are drawn, we use ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z}
to get a set of equations.
The forces acting directing on the hanging yoyo 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 yoyo is said to roll without slipping. That phrase means
that the angular acceleration of the yoyo 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.
ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z} 
T  mg = ma  rT = Iα_{cm} 
Since α_{cm} = a/r, we have two
simple equations
Eliminating T from the first equation yields,
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z}
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 f_{s}
_{MAX} unless it is explicitly stated.
ΣF_{x} = ma_{x}  ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z} 
mgsinθ  f_{s} = ma  N  mgcosθ = 0  Rf_{s} = I_{cyl}α_{cm} 
(a) The first equation is mgsinθ  ma = f_{s}. The third equation can be simplified by using the given value for I_{cyl} and by noting that α_{cm} = a/R. Then the third equation becomes
This can be substituted into the first equation to get
Solving for a yields
(b) We can use this result with f_{s} = ½ma, to get an expression for f_{s},
However, the maximum value of f_{s} is μ_{s}N.
The second equation gives N = mgcosθ, so
Using (1) and (2) to eliminate f_{s}, yields
Using the identity tanθ
= sinθ/cosθ,
we get
This is the angle at which the cylinder would start to slip as
it moved down the incline.
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z}
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 f_{s} _{MAX} unless it is explicitly stated.
ΣF_{x} = ma_{x}  ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z} 
mgsinθ  f_{s} = ma  N  mgcosθ = 0  Rf_{s} = I_{hoop}α_{cm} 
(a) The first equation is mgsinθ  ma = f_{s}. The third equation can be simplified by using the given value for I_{hoop} and by noting that α_{cm} = a/R. Then the third equation becomes
This can be substituted into the first equation to get
Solving for a yields
(b) We can use this result with f_{s} = ma, to get an expression for f_{s},
However, the maximum value of f_{s} is μ_{s}N. The second equation gives N = mgcosθ, so
Using (1) and (2) to eliminate f_{s}, yields
Using the identity tanθ
= sinθ/cosθ, we get
This is the angle at which the hoop would start to
slip as it moved down the incline.
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 ΣF_{x}
= ma_{x}, ΣF_{y}
= ma_{y}, and Στ_{z}
= I_{z}α_{z}
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
f_{s MAX} unless it is explicitly stated.
ΣF_{x} = ma_{x}  ΣF_{y} = ma_{y}  Στ_{z} = I_{z}α_{z} 
(m+¼M)gsinθ  f_{s} = (m+¼M)a  N  (m+¼M)gcosθ = 0  Rf_{s} = I_{disk}α_{cm} 
(a) The first equation is mgsinθ  ma = f_{s}. The third equation can be simplified by using I_{hoop} = ½mR^{2} and by noting that α_{cm} = a/R. Then the third equation becomes
This can be substituted into the first equation to get
Solving for a yields
(b) We can use this result with f_{s} = ½ma, to get an expression for f_{s},
However, the maximum value of f_{s} is
μ_{s}N. The second equation gives N
= (m+¼M)gcosθ, so
Using (1) and (2) to eliminate f_{s}, yields
Using the identity tanθ
= sinθ/cosθ, we get
This is the angle at which the wheels would start to slip as it moved down the incline.
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 f_{s} 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 F_{x} 
= ma_{x} 
Σ F_{y} 
= ma_{y} 
Σ τ_{cm} 
= I_{cmα cm} 
− Fcos(θ ) + f_{s} 
= − Ma 
N + Fsin(θ ) − Mg 
= 0 
Rf_{s} 
= Iα 
We also know that a = Rα and I = ½MR^{2}. We substitute these relationships into the torque equation
Rf_{s} = ½MR^{2}(a/R) 
(1) 
This yields an equation for f_{s},
f_{s} = ½Ma 
(2) 
We take this result and substitute it into the xcomponent 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 f_{s}, as required in part (b), by substituting (4) back into Eqn. (2);
f_{s} = (1/3)Fcos(θ ) 
(5) 
Part (c) asks us when the roller will slip. This occurs when f_{s} = f_{s}^{max}. Now we know f_{s}^{max} = μ_{s}N. We find N from the ycomponent equation, so
f_{s}^{max} = μ_{s}[Mg − Fsin(θ )] 
(6) 
Equating Eqns. (5) and (6) will tell us the value of F at which slipping occurs
(1/3)Fcos(θ )= μ_{s}[Mg − Fsin(θ )] 
(7) 
We get F by itself on the lefthand side
F[(1/3)cos(θ ) + μ_{s}sin(θ )] = μ_{s}Mg 
(8) 
So
(9) 
We know there is a normal and weight acting on the yoyo but these do not create torques as they operate on or through the Centre of Mass. There is nonmaximum 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 yoyo counterclockwise pushing the yoyo into the surface. The surface reacts by pushing back. The FBD looks like
Here I have guessed that the yoyo will roll backwards. Notice that my choice of a and α are consistent.
I apply Newton's Second Law
Σ F_{x } 
= ma_{x} 
Σ F_{y} 
= ma_{y} 
Σ τ_{cm} 
= I_{cm}a_{cm} 
T − f_{s} 
= − Ma 
N − Mg 
= 0 
rT − Rf_{s} 
= Iα 
We also know a = Rα .
The torque equation becomes can be solved for f_{s}
f_{s }= (r/R)T − (I/R^{2})a . 
(1) 
We substitute this into the xcomponent equation
T − [(r/R)T − (I/R^{2})a] = − Ma 
(2) 
We bring the term involving a from the left to the right and solve for a in terms of T,
T(Rr)/R = − (M + I/R^{2})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 f_{s} is positive, it must be have been chosen in the right direction.
Now as we see from Eqn. (5), as T increases so does f_{s}. We know f_{s} £ μ_{s}N or f_{s} £ μ_{s}Mg in our case when we make use of the ycomponent equation. Thus we have a limit on T
(6) 
This yields the result
(7) 
If T is any bigger, the yoyo will slip.
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 f_{k}
= μ_{k}N.
In this problem, N equals how hard the brake is pressed. Note that the
tension and the friction are tangential to the drum.
(a) W_{T} = (RT)Δφ = (0.5m)(100 N)(12π) = 1.88 × 10^{3} J.
(b) W_{f} = (rf_{k})Δφ = (0.35 m)(0.50 × 200 N)(12π) = 1.32 × 10^{3} J.
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 =
½MR^{2}. (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
nonconservative.
Since the problem involves a change is speed, we make
use
of the
Generalized WorkEnergy Theorem. Since there is only a change in
rotational speed,
The nonconservative force in this problem is tension, W_{NC}
= W_{T}. So we have
(a) The definition of work in rotational situations is W = τΔφ. Tension is always tangential to cylinders so τ_{T} = 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
(b) Then we find the final velocity from equation (1),
According to the table of Moments of Inertia, I =
½MR^{2}
for a solid cylinder. So
(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 WorkEnergy Theorem. Since there is only a change in rotational kinetic energy,
The nonconservative forces in this problem are the
tension and
the axle friction, W_{NC} = W_{T}
+ W_{f}. So we
have
The definition of work in rotational situations is W =
τΔφ. Tension is always tangential
to cylinders
so τ_{T}
= RT, again assuming the
rope pulls the cylinder clockwise. Thus the work done by the tension is
Using this result and equation (1), we can find the
work
done by
friction
Since W_{f} = τ_{f} Δφ, we find the frictional torque to be
the sign indicating that it is counterclockwise.
(c) After the rope comes off the cylinder, the only
force
acting
is friction so the generalized WorkEnergy Theorem becomes
where ω_{0} here is ω_{f} from the first part of the problem and the new ω_{f} = 0 since the drum comes to a stop. Again W_{f} = τ_{f}Δφ, where τ_{f} is the result from part (b). Therefore
(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
Rearranging for t yields,
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
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
Isolating v^{2} terms, we find v^{2}[m + I/R^{2}] = 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  I_{CM}  v 

solid sphere  ^{2}/_{5}MR^{2}  
solid cylinder  ½MR^{2}  
hollow sphere  ^{2}/_{3}MR^{2} 
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.
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
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 ½MR^{2}. Our equation becomes
Isolating v^{2} terms, we find v^{2}[½m + ¼m] = mgLsinθ  ½KL^{2}. Solving for v we get
(a) The problem involves a change in height and speed and has a spring, so we would apply the generalized WorkEnergy Theorem even if not directed to do,
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, W_{NC} = 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
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 = ½MR^{2}. 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,
Collecting the terms with v, and solving for v yields
(b) Recall from our discussions on kinematics that an object turns around when its velocity is zero. Setting equation (2) to zero
we see that the numerator is zero when
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
We see that the numerator to zero when
Solving this for h reveals that the object turns around when h = mg/k, halfway between the starting position and the turnaround point.
The problem involves changes in height, speed, and
rotation, so
we would apply the generalized WorkEnergy Theorem even if not directed
to do,
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, W_{NC} = 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
We are told that I_{disk} =
½M_{D}(R_{D})^{2}
and I_{hoop} = M_{H}(R_{H})^{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,
Collecting the terms with v, and solving for v yields
In the diagram block M_{1} is connected to M_{2} by a very light string running over three identical pulleys. The pulleys are disks with mass M_{p} and the rope does not slip. The coefficient of kinetic friction for the horizontal surface that M_{1} is on is μ_{k}. Find an expression for the speed v of block M_{2} in terms of the distance L that block M_{1} moves to the right. Your answer should be expressed in terms of L, M_{1}, M_{2}, M_{p}, μ_{k}, and g only.
The problem involves changes in height, speed, and rotation, so we would apply the generalized WorkEnergy Theorem even if not directed to do so,
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, W_{NC} = W_{f}. To find W_{f}, we need to the frictional force. To find a force, we draw a FBD and use Newton's Second Law.



ΣF_{x} = ma_{x}  ΣF_{y} = ma_{y}  
T  f_{k} = M_{1}a  N  M_{1}g = 0 
The second equation gives N = M_{1}g
and we know f_{k}
= μ_{k} N, so f_{k} =
μ_{k}M_{1}g.
Therefore, the work
done by friction is W_{f} = f_{k}
Δx
= μ _{k}M_{1}gΔx.
Next consider the change in energy of each object. M_{1}
increases its linear kinetic energy. M_{2} also
increases its
linear kinetic energy but loses gravitational potential energy. The
three identical pulleys increase their rotational kinetic energies.
Thus equation (1) becomes
Since M_{1} and M_{2}
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 I_{disk} = ½M_{P}R^{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,
Collecting the terms with v yields
Solving for v yields,
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 WorkEnergy Theorem even if not directed to do so,
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
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 I_{disk} = ½MR^{2}. 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
Collecting the terms with v yields
^{3}/_{4}Mv2 = FL – ½KL2 .
Solving for v yields,
v = [(4M – 2KL^{2}) / 3M]^{½} .
Questions? mikec@kwantlen.bc.ca