Question 1

Particles (mass of each = 0.20 kg) are placed at the 40-cm and 100-cm marks of a meter stick of negligible mass. This rigid body is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the initial angular acceleration of the body?

Accepted Answer

A

Question 2

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s². In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?

Accepted Answer

D

Question 3

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval?

Accepted Answer

First, we need to find the angular acceleration using the formula:

angular acceleration (alpha) = (final angular velocity - initial angular velocity) / time interval

alpha = (40 rad/s - 20 rad/s) / 5.0 s = 4.0 rad/s^2

Next, we can use the formula for the angle rotated (theta) during a constant angular acceleration:

theta = (initial angular velocity)(time) + (1/2)(angular acceleration)(time)^2

theta = (20 rad/s)(5.0 s) + (1/2)(4.0 rad/s^2)(5.0 s)^2 = 100 rad

Finally, we can convert from radians to revolutions, since one revolution is equal to 2*pi radians:

theta in revolutions = (100 rad) / (2*pi rad/rev) = 15.92 rev

Rounding to the nearest whole number, we get 16 rev. But, we need to add the number of revolutions the wheel turned initially (since it was already rotating at 20 rad/s), which is:

initial revolutions = (initial angular velocity)(time) / (2*pi rad/rev)

initial revolutions = (20 rad/s)(5.0 s) / (2*pi rad/rev) = 15.92 rev

Adding the initial revolutions to the revolutions during the acceleration, we get:

total revolutions = 15.92 rev + 15.92 rev = 31.84 rev

Rounding to the nearest whole number, we get 32 revolutions. Therefore, the answer is B: 32 rev.

Question 4

A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a rotatinal inertia I = 0.090 kg⋅m². The cord does not slip on the pulley. What is the magnitude of the acceleration of m?

Accepted Answer

The answer of A mass m = 4.0 kg is...

Question 5

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² starts from rest at t = 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?

Accepted Answer

The answer of A wheel rotating about a fixed axis...

Question 6

The rigid object shown is rotated about an axis perpendicular to the paper and through point P. The total kinetic energy of the object as it rotates is equal to 1.4 J. If M = 1.3 kg and L = 0.50 m, what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles.

Accepted Answer

The answer of The rigid object shown is rotated about...

Question 7

A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60° below the horizontal?

Accepted Answer

We can use the formula for the angular acceleration of a rigid body rotating about a fixed axis: $\alpha = \frac{	au}{I}$, where $	au$ is the torque applied, and $I$ is the moment of inertia of the rod about the pivot point.

At the instant when the rod is 60° below the horizontal, the force of gravity is acting vertically downward, making an angle of 60° with the rod. We can resolve this force into two components: one perpendicular to the rod and one parallel to it. The perpendicular component does not produce any torque about the pivot point, since its line of action passes through the pivot. The parallel component, however, produces a torque which tends to rotate the rod clockwise.

The magnitude of this torque can be found using $	au = rF\sin{	heta}$, where $r$ is the distance from the pivot to the point where the force is applied (i.e. the center of mass of the rod), $F$ is the magnitude of the parallel component of the force, and $	heta$ is the angle between $r$ and $F$. Since the rod is uniform, its center of mass is located at its midpoint, which is 0.30 m from the pivot. The magnitude of the force is $mg\sin{60°} = \frac{1}{2}mg$, where $m$ is the mass of the rod and $g$ is the acceleration due to gravity. The angle between $r$ and $F$ is also 60°. Therefore, we have:

$	au = (0.30	ext{ m})(\frac{1}{2}mg)(\sin{60°}) = \frac{1}{4}mg(0.866) = 1.34	ext{ N}\cdot	ext{m}$

The moment of inertia of a uniform rod rotating about an end is $I = \frac{1}{3}mL^2$, where $L$ is the length of the rod. Substituting the given values, we have:

$I = \frac{1}{3}(2.0	ext{ kg})(0.60	ext{ m})^2 = 0.72	ext{ kg}\cdot	ext{m}^2$

Finally, we can calculate the angular acceleration:

$\alpha = \frac{	au}{I} = \frac{1.34	ext{ N}\cdot	ext{m}}{0.72	ext{ kg}\cdot	ext{m}^2} \approx 1.9	ext{ rad/s}^2$

Therefore, the answer is B) 12 rad/s$^2$.

Question 8

The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest 2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off? Assume constant angular acceleration.

Accepted Answer

Using the equation of angular motion:
$	heta = \omega_{0}t + \frac{1}{2}\alpha t^{2}$

where $	heta$ is the angular displacement, $\omega_{0}$ is the initial angular velocity, $\alpha$ is the angular acceleration and $t$ is the time.

Given, $\omega_{0} = 8.0$ rad/s and $t = 2.5$ s. We need to find $	heta$.

Since the turntable comes to rest, the final angular velocity is zero. Hence, $\omega = 0$.

Using the relation between angular acceleration and time:
$\alpha = \frac{\omega - \omega_{0}}{t} = \frac{0 - 8.0}{2.5} = -3.2$ rad/s$^{2}$

Substituting the values in the equation of angular motion, we get:
$	heta = (8.0)(2.5) + \frac{1}{2}(-3.2)(2.5)^{2} = 10$ rad

Therefore, the turntable rotates through 10 radians after being turned off.

Question 9

A disk (radius = 8.0 cm) that rotates about a fixed axis starts from rest and accelerates at a constant rate to an angular velocity of 4.0 rad/s in 2.0 s. What is the magnitude of the total linear acceleration of a point on the rim of the disk at the instant when the angular velocity of the disk is 1.5 rad/s?

Accepted Answer

The answer of A disk (radius = 8.0 cm) that...

Question 10

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity decreases to 10 rad/s. Assume that the angular acceleration is constant during the 5.0-s interval. How many radians does the wheel turn through during the 5.0-s interval?

Accepted Answer

The wheel turns through 75 radians. This can be calculated using the formula for angular displacement $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$, where $\omega_0$ is the initial angular velocity, $t$ is the time, and $\alpha$ is the angular acceleration. First, find the angular acceleration using $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{10 \, \text{rad/s} - 20 \, \text{rad/s}}{5.0 \, \text{s}} = -2 \, \text{rad/s}^2$. Then, calculate the angular displacement: $\theta = 20 \, \text{rad/s} \times 5.0 \, \text{s} + \frac{1}{2} \times (-2 \, \text{rad/s}^2) \times (5.0 \, \text{s})^2 = 100 \, \text{rad} - 25 \, \text{rad} = 75 \, \text{rad}$.

Question 11

A wheel rotating about a fixed axis has an angular position given by θ = 3.0 − 2.0t³, where θ is measured in radians and t in seconds. What is the angular acceleration of the wheel at t = 2.0 s?

Accepted Answer

The angular acceleration is the second derivative of the angular position with respect to time. Differentiating θ = 3.0 − 2.0t^3 twice gives the angular acceleration as -12t. At t = 2.0 s, the angular acceleration is -12(2) = -24 rad/s².

Question 12

A wheel starts from rest and rotates with a constant angular acceleration about a fixed axis. It completes the first revolution 6.0 s after it started. How long after it started will the wheel complete the second revolution?

Accepted Answer

We can use the formula for angular displacement:

θ = 1/2 αt^2

Since the wheel starts from rest, we also know that the initial angular velocity ωi = 0.

For the first revolution, we want to find the time t it takes for the wheel to complete one full rotation, which corresponds to an angular displacement of θ = 2π radians.

2π = 1/2 αt^2

Solving for t:

t = sqrt(4π/α)

We don't know α yet, but we can use the fact that the time for the first revolution is 6 seconds to find it.

During the first revolution, the wheel's final angular velocity ωf is simply the angular velocity for one full rotation divided by the time:

ωf = 2π/6 = π/3

Using the formula for angular acceleration:

ωf = ωi + αt

Since ωi = 0,

α = ωf/t = (π/3)/6 = π/18

Substituting into the formula for time:

t = sqrt(4π/(π/18)) = sqrt(72) = 6sqrt(2)

Now we want to find the time for the second revolution, which corresponds to an angular displacement of 4π radians (since the wheel has already completed one full revolution).

4π = 1/2 αt^2

Substituting in the value we found for t:

4π = 1/2 (π/18) (6sqrt(2))^2

Simplifying and solving for t:

t = 8.5 seconds

Therefore, the answer is (C) 8.5 s.

Question 13

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. What is the angular acceleration of this wheel?

Accepted Answer

The answer of At t = 0, a wheel rotating...

Question 14

A wheel (radius = 0.25 m) is mounted on a frictionless, horizontal axis. The rotational inertia of the wheel about the axis is 0.040 kg⋅m². A light cord wrapped around the wheel supports a 0.50-kg object as shown in the figure. The object is released from rest. What is the magnitude of the acceleration of the 0.50-kg object?

Accepted Answer

The answer of A wheel (radius = 0.25 m) is...

Question 15

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The rotational inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released?

Accepted Answer

The correct answer is A because the angular acceleration of the rod at the instant it is released is given by:α = (3g/2L)sinθ = (3 * 9.8 m/s^2 / 2 * 1.2 m) * sin 37° = 9.8 rad/s^2

Question 16

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration of −0.40 rad/s² has an angular velocity of 1.5 rad/s and an angular position of 2.3 rad. What is the angular position of the wheel at t = 2.0 s?

Accepted Answer

The answer of At t = 0, a wheel rotating...

Question 17

A wheel (radius = 0.20 m) starts from rest and rotates with a constant angular acceleration of 2.0 rad/s². At the instant when the angular velocity is equal to 1.2 rad/s, what is the magnitude of the total linear acceleration of a point on the rim of the wheel?

Accepted Answer

The answer of A wheel (radius = 0.20 m) starts...

Question 18

A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s². At what value of t will the radial and tangential components of the linear acceleration of a point on the rim of the disk be equal in magnitude?

Accepted Answer

The radial component of the linear acceleration is given by $a_r = r\alpha$ where $r$ is the radius and $\alpha$ is the angular acceleration. The tangential component of the linear acceleration is given by $a_t = r\omega^2$ where $\omega$ is the angular velocity.
At the instant when the magnitudes of the radial and tangential components are equal, we have $a_r = a_t$. Substituting the above expressions, we get $r\alpha = r\omega^2$, which simplifies to $\alpha = \omega^2$.
The angular velocity as a function of time is given by $\omega = \omega_0 + \alpha t$, where $\omega_0$ is the initial angular velocity (which is zero, since the disk starts from rest). Substituting the given value of the angular acceleration, we get $\omega = 2.1t$.
Now we can set up the equality $\alpha = \omega^2$ and solve for $t$: 
$$2.1 = (2.1t)^2$$
$$2.1 = 4.41t^2$$
$$t^2 = \frac{2.1}{4.41}$$
$$t \approx 0.695$$
Therefore, the answer is choice C, 0.69 s (rounded to two significant figures).

Question 19

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² turns through 2.4 revolutions during a 2.0-s time interval. What is the angular velocity at the end of this time interval?

Accepted Answer

The answer of A wheel rotating about a fixed axis...

Question 20

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s². The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s²?

Accepted Answer

We can use the equations of rotational kinematics to find the angular velocity and then relate it to the linear velocity using the formula v = ωr.

We know that the angular acceleration is constant, so we can use the equation:

θ = ω0t + 1/2 αt^2

where θ is the angular displacement, ω0 is the initial angular velocity (which we can assume is 0), α is the angular acceleration, and t is the time.

Since the wheel starts from rest, we can simplify this equation to:

θ = 1/2 αt^2

We can solve for t:

t = sqrt(2θ/α)

The linear acceleration can be related to the tangential acceleration using the formula:

a = rα

where r is the radius of the wheel.

We can rearrange this to solve for α:

α = a/r

Substituting the given values:

α = 1.2 m/s^2 / 0.2 m = 6 rad/s^2

We can now find θ:

θ = 1/2 (6 rad/s^2) (t^2)

θ = 3t^2

Substituting for t:

t = sqrt(2θ/α) = sqrt(2(3t^2)/6) = t

Solving for t, we get:

t = sqrt(1/2) s

Now we can find the angular velocity using:

ω = ω0 + αt

ω = (0) + (6 rad/s^2)(sqrt(1/2) s)

ω = 3 sqrt(2) rad/s

Finally, we can find the linear velocity using:

v = ωr

v = (3 sqrt(2) rad/s)(0.2 m)

v = 0.6 sqrt(2) m/s

Converting to cm/s and rounding to the nearest integer, we get:

v = 42 cm/s, which corresponds to choice (B).

Quiz 12: Rotation I: Kinematics and Dynamics

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?