Question 1

A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kg  and a radius of 12.5 cm. A mass of 423 g is attached to the free end of the string. With the string vertical and taut, the mass is released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate. What is the speed of the mass after it has fallen through 1.25 m?

Accepted Answer

The answer of A string is wrapped tightly around a...

Question 2

FIGURE 10-2 
-Figure 10-2 illustrates a simplified roller bearing. The inner cylinder has a radius of 1.0 cm and is stationary. The outer hollow cylinder has a radius of 1.2 cm and is rotating at 10 rpm. Between the two cylinders are several small cylinders with a radius of 0.10 cm, which roll without slipping on both the inner and outer cylinders. Only one of these cylinders is shown in the figure. What is the angular speed of the small cylinders?

Accepted Answer

The small cylinders roll without slipping, so their linear speed at the point of contact with both the inner and outer cylinders must be the same. The outer cylinder's linear speed is 2π(1.2 cm)(10 rpm) = 24π cm/min. The small cylinders have a radius of 0.10 cm, so their angular speed is (24π cm/min) / (0.10 cm) = 240 rpm. However, since they are in contact with both cylinders, the effective angular speed is half of this, 120 rpm, but relative to the stationary inner cylinder, it is 60 rpm.

Question 3

A pencil, 15.7 cm long, is released from a vertical position with the eraser end resting on a table. The eraser does not slip. Treat the pencil like a uniform rod. What is the angular speed of the pencil when it makes a 30.0° angle with the vertical?

Accepted Answer

The answer of A pencil, 15.7 cm long, is released...

Question 4

A 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?

Accepted Answer

When a solid sphere rolls without slipping, its kinetic energy is given by the sum of its translational and rotational kinetic energies:

KE = 1/2 mv^2 + 1/2 Iω^2

where m is the mass of the sphere, v is its linear speed, I is its moment of inertia, and ω is its angular velocity.

In this case, because the sphere is rolling without slipping, we have

v = Rω

where R is the radius of the sphere. We can use this relationship to eliminate ω in favor of v in the equation above:

KE = 1/2 mv^2 + 1/2 (2/5 mR^2)(v/R)^2
   = 1/2 mv^2 + 1/5 mv^2
   = 7/10 mv^2

Note that we've used the fact that for a solid sphere, I = 2/5 mR^2.

Now, as the sphere rolls up the ramp, it gains gravitational potential energy:

PE = mgh

where h is the height it has climbed. Conservation of energy tells us that the increase in the sphere's potential energy must come at the expense of its kinetic energy, so we have

mgh = KEi - KEf

where KEi is the sphere's initial kinetic energy (when it is rolling without slipping along the horizontal surface) and KEf is its final kinetic energy (when it has rolled up the ramp).

At the start of the ramp, the sphere has only translational kinetic energy, so

KEi = 1/2 mv^2

At the top of the ramp, the sphere has both translational and rotational kinetic energy, but because it is rolling without slipping, we have

v = ωR

and

KEf = 1/2 mv^2 + 1/2 Iω^2
    = 1/2 mv^2 + 1/2 (2/5 mR^2)(v/R)^2
    = 1/2 mv^2 + 1/5 mv^2
    = 7/10 mv^2

(where we've used the same calculations as before to find the rotational kinetic energy).

Putting it all together:

mgh = 1/2 mv^2 - 7/10 mv^2

Solving for v:

v = sqrt[(2gh)/7]
  = sqrt[(2 * 9.81 m/s^2 * 3.00 m)/7]
  ≈ 3.53 m/s

So the answer is E.

Question 5

A hoop 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 hoop after it has rolled 3.00 m up the ramp?

Accepted Answer

First, we need to find the initial energy of the hoop. Since it is rolling without slipping, we can use the formula $E=\frac{1}{2}(I_{cm}+mR^2)\omega^2+\frac{1}{2}mv^2$, where $I_{cm}$ is the moment of inertia of the hoop about its center of mass, $m$ is the mass of the hoop, $R$ is the radius of the hoop, $\omega$ is the angular velocity of the hoop (which is related to the linear speed $v$ by $\omega=\frac{v}{R}$), and $v$ is the speed of the hoop.

For a uniform hoop, $I_{cm}=\frac{1}{2}mR^2$, so the formula simplifies to $E=\frac{3}{4}mv^2$. Plugging in $v=5.50	ext{ m/s}$, we find that the initial energy is $E_i=\frac{3}{4}m(5.50	ext{ m/s})^2=\frac{75}{4} m	ext{ m}^2/	ext{s}^2$.

As the hoop rolls up the ramp, its potential energy increases and its kinetic energy decreases due to the work done by gravity. When it reaches the top of the ramp, all of its initial kinetic energy will have been converted into potential energy, so we can set the final kinetic energy equal to zero and solve for the final speed:

\begin{align*}
E_i&=E_f\
\frac{3}{4}mv_i^2&=mg(h+\frac{1}{2}R\sin	heta)\
v_f&=\sqrt{\frac{4}{3}g(h+\frac{1}{2}R\sin	heta)}\
&=\sqrt{\frac{4}{3}\cdot9.81	ext{ m/s}^2\cdot(3.00	ext{ m}+\frac{1}{2}(0.20	ext{ m})\sin(25.0^\circ))}\
&=\boxed{	extbf{(A) }4.22	ext{ m/s}}
\end{align*}

Question 6

A pencil, 15.7 cm long, is released from a vertical position with the eraser end resting on a table. The eraser does not slip. Treat the pencil like a uniform rod. What is the angular speed of the pencil just before it hits the table?

Accepted Answer

The correct answer is A because the angular speed of the pencil just before it hits the table is 17.2 rad/s. This can be calculated using the formula:ω = √(2gh/L)where:ω is the angular speed in radians per secondg is the acceleration due to gravity (9.8 m/s^2)h is the height from which the pencil is released (15.7 cm = 0.157 m)L is the length of the pencil (15.7 cm = 0.157 m)Plugging in the values, we get:ω = √(2 * 9.8 m/s^2 * 0.157 m / 0.157 m) = 17.2 rad/s

Question 7

A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kg m² and a radius of 12.5 cm. The string is pulled away from the pulley with a constant force of 5.00 N. As the string unwinds the pulley begins to rotate. What is the speed of the string after it has unwound 1.25 m?

Accepted Answer

First, we need to calculate the torque applied to the pulley by the force pulling the string. The torque is given by:

τ = rF = (0.125 m)(5.00 N) = 0.625 Nm

Next, we need to find the angular acceleration of the pulley using the moment of inertia:

τ = Iα

α = τ/I = 0.625 Nm / 0.0352 kg m2 = 17.76 rad/s2

The final angular velocity of the pulley can be found using the kinematic equation:

ωf^2 = ωi^2 + 2αΔθ

Since the pulley rotates in the same direction as the unwinding string, the angle of rotation is given by the distance the string has unwound divided by the pulley radius:

Δθ = Δx / r = 1.25 m / 0.125 m = 10 rad

Plugging in the values, we get:

ωf^2 = 0 + 2(17.76 rad/s2)(10 rad)
ωf^2 = 355.2
ωf = 18.85 rad/s

Finally, we convert the angular velocity to linear velocity by multiplying by the pulley radius:

v = ωf r = (18.85 rad/s)(0.125 m) = 2.36 m/s

Therefore, the answer is B.

Question 8

A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 3.5 rad/s. What is its kinetic energy?

Accepted Answer

The answer of A solid cylinder with a radius of...

Question 9

A solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25.0° with the horizontal. What is the speed of the disk after it has rolled 3.00 m, measured along the plane?

Accepted Answer

Let's use conservation of energy here. The initial potential energy of the disk is converted to both translational and rotational kinetic energy. Therefore we can write:
Mgh = (1/2)MV^2 + (1/2)Iω^2
where M is the mass of the disk, h is the height of the inclined plane, V is the velocity of the disk, I is the moment of inertia of the disk, and ω is the angular velocity of the disk.
We know that the disk is rolling without slipping. Therefore we can relate the linear velocity to the angular velocity as V = Rω, where R is the radius of the disk.
Also, for a solid disk, the moment of inertia is (1/2)MR^2.
Substituting all these values into the energy equation and rearranging, we get:
V = sqrt(2gh/3) = 4.07 m/s
Therefore, the answer is (C).

Question 10

An object is made up of three masses connected by massless rods of fixed length. Mass A is located at (30.0 cm, 0 cm) and has a mass of 250 grams, mass B is located at (0 cm, 30.0 cm) and has a mass of 350 grams, mass C is located at (-30.0 cm, 0 cm) and has a mass of 450 grams. What is the moment of inertia of this object about an axis perpendicular to the x-y plane and passing through the origin?

Accepted Answer

The moment of inertia of the object about an axis perpendicular to the x-y plane and passing through the origin is given by:$$I = mA^2 + mB^2 + mC^2$$where m is the mass of each object and A, B, and C are the distances from the axis to the center of mass of each object.In this case, we have:$$mA = 250 g = 0.25 kg$$$$A = 30.0 cm = 0.3 m$$$$mB = 350 g = 0.35 kg$$$$B = 30.0 cm = 0.3 m$$$$mC = 450 g = 0.45 kg$$$$C = 30.0 cm = 0.3 m$$So, the moment of inertia is:$$I = (0.25 kg)(0.3 m)^2 + (0.35 kg)(0.3 m)^2 + (0.45 kg)(0.3 m)^2$$$$I = 0.0945 kg m^2$$Therefore, the correct answer is A.

Question 11

In a part of a printing press, a roller with a radius of 10 cm, rolls without slipping on the outside of a fixed cylinder with a radius of 30 cm. There is a connecting rod that connects the center of the cylinder to the axle of the roller. If the angular speed of the roller is 60 rad/s, what is the angular speed of the connecting rod?

Accepted Answer

First, we need to find the linear speed of the roller by using the formula: 
v = ωr 
where ω is the angular speed in radians per second, and r is the radius in meters.

v = (60 rad/s)(0.1 m) = 6 m/s

Next, we can use the fact that the roller is rolling without slipping to relate the linear speeds of the roller and the cylinder:

v_roller = v_cylinder

ω_roller r_roller = ω_cylinder r_cylinder

ω_cylinder = (ω_roller r_roller) / r_cylinder

ω_cylinder = (60 rad/s)(0.1 m) / 0.3 m = 20 rad/s

Therefore, the angular speed of the connecting rod is also 20 rad/s (since it is connected to the center of the cylinder).

The correct answer is B.

Question 12

The moment of inertia of a uniform rod (about its center) is given by I = ML²/12. What is the kinetic energy of a 120-cm rod with a mass of 450 g rotating about its center at 3.60 rad/s?

Accepted Answer

The kinetic energy of a rotating rod is given by K = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. Substituting the given values, we get:K = (1/2)(ML^2/12)(3.60 rad/s)^2 = (1/2)(0.450 kg)(1.20 m)^2(3.60 rad/s)^2 = 0.350 J

Question 13

An Atwood machine has a mass of 3.50 kg connected by a light string to a mass of 6.00 kg over a pulley with a moment of inertia of 0.0352 kg m² and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m?

Accepted Answer

First, we need to find the net force acting on the system. Using Newton's second law, 
$F_{net} = (m_2 - m_1)g$, where $m_1$ = 3.50 kg and $m_2$ = 6.00 kg.
$F_{net} = (6.00 	ext{ kg} - 3.50 	ext{ kg})\cdot 9.81 	ext{ m/s}^2 = 23.4 	ext{ N}$

Next, we can use the torque equation, $	au = I\alpha$, to find the angular acceleration of the pulley. Since the pulley is a disk, $I = \frac{1}{2}MR^2$.
$	au = Fr = (m_2 - m_1)g\cdot r = 23.4	ext{ N}\cdot 0.125	ext{ m} = 2.93 	ext{ N}\cdot	ext{m}$
$I\alpha = 	au \Rightarrow \alpha = \frac{	au}{I} = \frac{2\cdot 2.93	ext{ N}\cdot	ext{m}}{0.0352	ext{ kg}\cdot	ext{m}^2} = 166.2 	ext{ rad/s}^2$

Now we can apply kinematic equations to find the speed of the masses after they have moved a distance of 1.25 m. Since the pulley does not slip, the distance travelled by the masses is equal to the distance travelled by the edge of the pulley, which is $1.25m\cdot\frac{2}{\pi} \approx 0.795m$.
Using the kinematic equation $v^2 = v_0^2 + 2a\Delta x$, where $v_0 = 0$ and $\Delta x = 0.795m$ (for the edge of the pulley):
$v = \sqrt{2a\Delta x} = \sqrt{2\cdot 166.2 	ext{ rad/s}^2\cdot 0.795	ext{ m}} \approx 2.28 	ext{ m/s}$
Therefore, the answer is (B) 2.28 m/s.

Question 14

FIGURE 10-2 
-Figure 10-2 illustrates a simplified roller bearing. The outer hollow cylinder has a radius of 1.2 cm and is stationary. The inner cylinder has a radius of 1.0 cm and is rotating at 10 rpm. Between the two cylinders are several small cylinders with a radius of 0.10 cm, which roll without slipping on both the inner and outer cylinders. Only one of these cylinders is shown in the figure. What is the angular speed of the small cylinders?

Accepted Answer

The answer of FIGURE 10-2 
-Figure 10-2 illustrates a simplified...

Question 15

A uniform ball with a mass of 125 g is rolling without slipping along the horizontal surface of a table with a speed of 4.50 m/s when it rolls off the edge and it falls towards the floor, 1.10 m below. What is the rotational kinetic energy of the ball just before it hits the floor?

Accepted Answer

The rotational kinetic energy of a rolling ball can be calculated using the formula $ KE_{rot} = \frac{1}{2} I \omega^2 $. For a solid sphere, $ I = \frac{2}{5} m r^2 $ and $ \omega = \frac{v}{r} $. Substituting these into the formula gives $ KE_{rot} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{5} m v^2 $. Using $ m = 0.125 \, \text{kg} $ and $ v = 4.50 \, \text{m/s} $, we find $ KE_{rot} = \frac{1}{5} \times 0.125 \times (4.50)^2 = 0.506 \, \text{J} $. The radius cancels out, so it is not needed.

Question 16

Two wheels with fixed centers are in contact with each other and rotate without slipping. Wheel A has a radius of 12.0 cm and is rotating with an angular speed of 35.0 rad/s. Wheel B has a radius of 17.0 cm. What is the angular speed of wheel B?

Accepted Answer

The angular velocity of the two wheels is the same since they are in contact with each other and rotate without slipping. 
ω = ω₁ = ω₂
Therefore, we can use the equation:
v = rω
where v is the tangential velocity and r is the radius. We can assume that the point of contact between the two wheels is not moving (i.e. v = 0) and solve for ω₂:
r₂ω₂ = r₁ω₁
17 cm * ω₂ = 12 cm * 35 rad/s
ω₂ = (12 cm * 35 rad/s) / 17 cm
ω₂ = 24.7 rad/s

Question 17

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around the inner core. The spool is free to roll without slipping on a horizontal surface. If the string unwinds horizontally from the bottom of the core with a constant speed of 25.0 cm/s, what is the speed of the spool?

Accepted Answer

The answer of A spool whose inner core has a...

Question 18

FIGURE 10-2 
-Figure 10-2 illustrates a simplified roller bearing. The outer hollow cylinder has a radius of 1.20 cm and is stationary. The inner cylinder has a radius of 1.00 cm and is rotating at 12.0 rpm. Between the two cylinders are several small cylinders with a radius of 0.100 cm, which roll without slipping on both the inner and outer cylinders. Only one of these cylinders is shown in the figure. How long does it take a small cylinder to complete a full revolution around the inner cylinder?

Accepted Answer

The small cylinder rolls without slipping, so its linear speed is equal to the difference in circumferences of the inner and outer cylinders. The time for one revolution is the circumference of the path divided by this speed.

Question 19

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around the inner core. The spool is free to roll without slipping on a horizontal surface. If the string unwinds horizontally from the top of the core with a constant speed of 29.0 cm/s, what is the speed of the spool?

Accepted Answer

The correct answer is A because the speed of the spool is equal to the speed of the string unwinding from the core.

Question 20

A solid disk is rolling without slipping along a horizontal surface with a speed of 4.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the disk after it has rolled 3.00 m up the ramp?

Accepted Answer

The answer of A solid disk is rolling without slipping...

Quiz 10: Rotational Kinematics and Energy