Homework 8 - Rotational Kinematics and Dynamics

question 1:

In the figure, a given force F is applied to a rod in several different ways. In which case is the torque due to F about the pivot P greatest?

a) 1

b) 5

c) 3

d) 4

e) 2

a) 1

explanation for question 1 (In the figure, a given force F is applied to a rod in several different ways. In which case is the torque due to F about the pivot P greatest?)

1

torque is greatest when the force applied is as far away from the axis of rotation as possible, and perpendicular to the line of action.

This is because torque is calculated by force x lever arm, or radius x force x sin(theta). Sine is greatest at 90*.

question 2:

If you deform an object, you do not change its mass but you may change its moment of inertia and the location of its center of mass.

a) true

b) false

a) true

explanation for question 2 (If you deform an object, you do not change its mass but you may change its moment of inertia and the location of its center of mass.)

true

Think of punching or molding clay. From the same chunk of clay, you could make a round ball or a deformed blob. The round ball and deformed blob would balance differently, even though they have the same mass. If the mass of an object is changed, it should be considered a different object.

question 3:

If two objects have the same moment of inertia, they must have the same mass.

a) true

b) false

b) false

explanation for question 3 (If two objects have the same moment of inertia, they must have the same mass.)

false

moment of inertia depends on both mass and radius of an object. Two objects could have the same moment of inertia, but different masses. The radii would be different, making the I values equal.

For example, two objects might both have a moment of inertia of 20 kg.m². The first object’s mass is 3 kg and the second object’s mass is 5 kg. The first object’s radius would be roughly 2.58 m, while the second object’s radius would be 2 m.

To check: I1 = (3)(2.58)² = 19.97 ~ 20 kg.m² and I2 = (5)(2)² = 20 kg.m². The two objects have the same moment of inertia, even though they have different masses.

question 3:

A cylinder and a sphere, both solid and uniform and having the same mass and diameter, roll without slipping down the same ramp starting from rest. Both of them will reach the ground at the same time.

a) true

b) false

b) false

explanation for question 4 (A cylinder and a sphere, both solid and uniform and having the same mass and diameter, roll without slipping down the same ramp starting from rest. Both of them will reach the ground at the same time.)

false

The two shapes have different moments of inertia, as provided in the formula sheet. If their moments of inertia are different, then their kinetic energies are different, and thus they will reach the ground at different times.

question 5:

A string is wrapped around a pulley of radius 0.5 m and moment of inertia 2 kg × m².  If the string is pulled with a force F, the resulting angular acceleration of the pulley is 2 rad/s².  Determine the magnitude of the force. 

a) 8 N

b) 0.2 N

c) 2 N

d) 0.8 N

a) 8 N

explanation for question 5 (A string is wrapped around a pulley of radius 0.5 m and moment of inertia 2 kg × m².  If the string is pulled with a force F, the resulting angular acceleration of the pulley is 2 rad/s².  Determine the magnitude of the force.)

8 N

This can be solved using the equations for torque. Torque can either be solved as force x lever arm or radius, or moment of inertia x angular acceleration.

τ = Fl = Fr (note, l and r both represent length of lever arm, or radius from axis of rotation)

τ = Iα

Thus, Fr = Iα. r, I, and α are known, so F can be isolated.

F = Iα/r → (2)(2)/(0.5) = 4/0.5 = 8 N

question 6:

The drawing shows a system of objects consisting of three small balls connected by massless rods. The axis of rotation is at m_1, and it is perpendicular to the page, as shown.  The force of magnitude F of magnitude 424 N  is applied to the ball (see the drawing). The masses of the balls are m₁ = 9.0 kg, m₂ = 6.0 kg, and  m₃ = 7.0 kg. 

The moment of inertia about the given axis of rotation is:

a) 229 kg.m²

b) 175 kg.m²

c) 53 kg.m²

d) 18 kg.m²

a) 229 kg.m²

explanation for question 6 (The drawing shows a system of objects consisting of three small balls connected by massless rods. The axis of rotation is at m_1, and it is perpendicular to the page, as shown.  The force of magnitude F of magnitude 424 N  is applied to the ball (see the drawing). The masses of the balls are m₁ = 9.0 kg, m₂ = 6.0 kg, and  m₃ = 7.0 kg. 

The moment of inertia about the given axis of rotation is:)

229 kg.m²

This problem uses the equation for moment of inertia, I = Σm_ir_i².

For this problem, I = (m₁r₁²) + (m₂r₂²) + (m₃r₃²)

[(9.0)(0)²] + [(6.0)(3)²] + [(7.0)(5)²] = 229 kg.m²

(the distance between m₂ and m₃, although provided, is not needed for this question. m₁ has no radius because it is the axis of rotation)

question 7:

A block of mass 2.0 kg is hanging from a massless cord that is wrapped around a pulley (I =1.1 x10^-3 kg.m²), as the drawing shows. Initially, the pulley is prevented from rotating, and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.040 m during the block's descent. 

The angular acceleration of the pulley is:

a) 6.22 rad/s

b) 38.15 rad/s

c) 17.52 rad/s

d) 182.3 rad/s

d) 182.3 rad/s

explanation for question 7 (A block of mass 2.0 kg is hanging from a massless cord that is wrapped around a pulley (I =1.1 x10^-3 kg.m²), as the drawing shows. Initially, the pulley is prevented from rotating, and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.040 m during the block's descent. 

The angular acceleration of the pulley is:)

182.3 rad/s²

To solve, it should be known that F = ma = mg - T, where m = mass, a = linear acceleration, g = gravitational force, and T = tension.

a = rα, so F can be written as F = mrα = mg - T.

Remember that tension is a force.

Force can be connected to moment of inertia through the two equations for torque. Since a value for force is not given, we want to eliminate it. τ = Fr and also τ = Iα. When set equal so that Fr = Iα, F can be isolated so F = Iα/r. Let F represent the tension force.

Going back up to our earlier equations, F = mrα = mg - Iα/r. Put α on the same side of the equation. mg = mrα + Iα/r.

Isolate α so that mg = α(mr + I/r), and solve.

(2.0)(9.8) = α((2.0)(0.04) + (1.1 × 10^-3)/(0.04))

19.6 = α(0.08 + 0.0275) → 19.6 = α(0.1075)

α = 182.33 rad/s²

question 8:

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the door's surface. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s^2. What is the moment of inertia of the door about the hinges?

a) 2.74 kg.m²

b) 4.28 kg.m²

c) 7.52 kg.m²

d) 1.88 kg.m²

e) 2.00 kg.m²

e) 2.00 kg.m²

explanation for question 8 (A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the door's surface. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s^2. What is the moment of inertia of the door about the hinges?)

2.00 kg.m²

moment of inertia for this problem can be solved with the equations for torque. Do not use the equation for moment of inertia involving mass and radius.

τ = Fr = Iα. F, r, and α are all known. I can be isolated so that I = Fr/α.

I = (5.0)(0.8)/(2.0) = 2 kg.m²

(note: I = m_ir_i² is not used because the mass of the door and doorknob are not given. The mass provided is for the person, not the door. This question does not involve the person at all.)

(note 2: r also stands for l, lever arm. I just used r so that moment of inertia I and lever arm l would not get mixed up)

question 9:

A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope wound on the wheel and attached to it. The wheel is released from rest, and the block descends 1.5 m in 2.00 s without slipping the rope. The tension in the rope during the block's descent is 20 N. What is the moment of inertia of the wheel?

a) 3.9 kg.m²

b) 4.3 kg.m²

c) 3.7 kg.m²

d) 4.1 kg.m²

e) 3.5 kg.m²

b) 4.3 kg.m²

explanation for question 9 (A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope wound on the wheel and attached to it. The wheel is released from rest, and the block descends 1.5 m in 2.00 s without slipping the rope. The tension in the rope during the block's descent is 20 N. What is the moment of inertia of the wheel?)

4.3 kg.m²

idk why

question 10:

A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not turning. A constant force of 50 N is applied to the string, which does not slip, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s , what is the moment of inertia of the pulley?

a) 0.17 kg.m²

b) 14 kg.m²

c) 0.20 kg.m²

d) 17 kg.m²

c) 0.20 kg.m²

explanation for question 10 (A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not turning. A constant force of 50 N is applied to the string, which does not slip, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s , what is the moment of inertia of the pulley?)

0.20 kg.m²

idk why

question 11:

A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?

a) 4.01 m/s

b) 2.16 m/s

c) 8.02 m/s

d) 3.53 m/s

e) 1.91 m/s

c) 3.53 m/s

explanation for question 11 (A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?)

3.53 m/s

idk why

question 12:

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the disk's angular velocity is 5.35 rad/s at the bottom, what is the height of the inclined plane?

a) 4.21 m

b) 6.73 m

c) 4.94 m

d) 5.61 m

d) 5.61 m

explanation for question 12 (A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the disk's angular velocity is 5.35 rad/s at the bottom, what is the height of the inclined plane?)

5.61 m

idk why

question 13:

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg • m2 is rotating freely with an angular speed of 1.0 rad/s. Two people, each having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on?

a) 0.60 rad/s

b) 0.20 rad/s

c) 0.80 rad/s

d) 0.67 rad/s

e) 0.40 rad/s

e) 0.40 rad/s

explanation for question 13 (A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg • m2 is rotating freely with an angular speed of 1.0 rad/s. Two people, each having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on?)

0.40 rad/s

idk why

question 14:

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg.m². If the arms are pulled in, so the moment of inertia decreases to 1.8 kg,m², what is the final angular speed?

a) 6.25 rad/s

b) 0.81 rad/s

c) 2.25 rad/s

d) 4.60 rad/s

e) 1.76 rad/s

a) 6.25 rad/s

explanation for question 14 (A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg.m². If the arms are pulled in, so the moment of inertia decreases to 1.8 kg,m², what is the final angular speed?)

6.25 rad/s

idk why