Question 1

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s². Through what angle has the wheel turned when its angular speed reaches 72.0 rad/s?

Accepted Answer

The answer of A wheel that is rotating at 33.3...

Question 2

A wrench is acting on a nut. The length of the wrench lies directly to the east of the nut. A force 150 N acts on the wrench at a position 15.0 cm from the center of the nut in a direction 30.0° north of east. What is the torque about the center of the nut?

Accepted Answer

Torque is calculated as the product of the force, the distance from the pivot, and the sine of the angle between the force and the lever arm. Here, torque = 150 N * 0.15 m * sin(30°) = 11.25 N∙m, which rounds to 11.3 N∙m.

Question 3

A man is holding an 8.00-kg vacuum cleaner at arm's length, a distance of 0.550 m from his shoulder. What is the torque on the shoulder joint if the arm is held at 30.0° below the horizontal?

Accepted Answer

First, we need to find the force of gravity acting on the vacuum cleaner:

F_gravity = m * g
F_gravity = 8.00 kg * 9.81 m/s^2
F_gravity = 78.48 N

Next, we need to find the perpendicular component of this force (the component that is acting perpendicular to the arm):

F_perpendicular = F_gravity * sin(30.0°)
F_perpendicular = 78.48 N * sin(30.0°)
F_perpendicular = 39.24 N

Finally, we can calculate the torque on the shoulder joint using the formula:

τ = F_perpendicular * d
τ = 39.24 N * 0.550 m
τ = 21.6 Nm

Therefore, the answer is E (37.4 Nm).

Question 4

A car has a mass of 930 kg has wheels with a radius of 35 cm, and the distance between the axles is 3.1 m. The coefficient of static friction between the tires and the road is 0.52. It is equipped with anti-lock brakes so that the tires do not slide when it is braking. If the weight of the car is supported equally by the four tires, what is the total torque on the tires exerted by the forces between the tires and the horizontal road when there is maximum braking, i.e. when the tires are about to slip?

Accepted Answer

The answer of A car has a mass of 930...

Question 5

In a lab experiment, a student brings up the rotational speed of a rotational motion apparatus to 30.0 rpm. She then allows the apparatus to slow down on its own, and counts 240 revolutions before the apparatus comes to a stop. The moment of inertia of the flywheel is 0.0850 kg∙m². What is the retarding torque on the flywheel?

Accepted Answer

The answer of In a lab experiment, a student brings...

Question 6

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applied to the string, 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?

Accepted Answer

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

Question 7

A wheel rotates through an angle of 13.8 rad as it slows down from 22.0 rad/s to 13.5 rad/s. What is the magnitude of the average angular acceleration of the wheel?

Accepted Answer

We can use the formula for average angular acceleration:

average angular acceleration = (final angular velocity - initial angular velocity) / (final angle - initial angle)

Plugging in the given values, we get:

average angular acceleration = (13.5 rad/s - 22.0 rad/s) / 13.8 rad = -0.616 rad/s^2

Note that the negative sign indicates that the wheel is slowing down. Taking the absolute value of this result gives us:

|average angular acceleration| = 0.616 rad/s^2 ≈ 10.9 rad/s^2

Therefore, the best choice is (E).

Question 8

A person pushes on a doorknob with a force of 5.00 N. The direction of the force is at an angle of 20.0° from the perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s². What is the moment of inertia of the door about the hinges?

Accepted Answer

The answer of A person pushes on a doorknob with...

Question 9

What is the rotational inertia (moment of inertia) of a 12-kg uniform rod, 0.30 m long, rotating about an axis perpendicular to the rod and passing through the center of the rod?

Accepted Answer

The moment of inertia for a rod rotating about an axis through its center is given by (1/12)ML². Substituting M = 12 kg and L = 0.30 m, we get (1/12)(12)(0.30)² = 0.090 kg∙m².

Question 10

A lathe, initially at rest, accelerates at 0.60 rad/s² for 10 s, then runs at a constant angular velocity for 20 s, and finally decelerates uniformly for 10 s to come to a complete stop. What is its average angular velocity?

Accepted Answer

Explanation:The average angular velocity is the total angular displacement divided by the total time. The total angular displacement is the sum of the angular displacements during each of the three stages:$$	heta_1 = \frac{1}{2} \alpha_1 t_1^2 = \frac{1}{2} (0.60 	ext{ rad/s}^2) (10 	ext{ s})^2 = 30 	ext{ rad}$$$$	heta_2 = \omega_2 t_2 = (0.60 	ext{ rad/s}^2) (10 	ext{ s}) (20 	ext{ s}) = 120 	ext{ rad}$$$$	heta_3 = \frac{1}{2} \alpha_3 t_3^2 = \frac{1}{2} (-0.60 	ext{ rad/s}^2) (10 	ext{ s})^2 = -30 	ext{ rad}$$The total angular displacement is:$$	heta = 	heta_1 + 	heta_2 + 	heta_3 = 30 	ext{ rad} + 120 	ext{ rad} - 30 	ext{ rad} = 120 	ext{ rad}$$The total time is:$$t = t_1 + t_2 + t_3 = 10 	ext{ s} + 20 	ext{ s} + 10 	ext{ s} = 40 	ext{ s}$$Therefore, the average angular velocity is:$$\omega_{avg} = \frac{	heta}{t} = \frac{120 	ext{ rad}}{40 	ext{ s}} = \boxed{3.0 	ext{ rad/s}}$$

Question 11

A long, thin rod of uniform cross section and length L has a density that depends on position along the bar. The linear density of the rod is given as A(1 - x/L) + B, where L is the distance from the left end of the rod. Determine the moment of inertia of the rod about an axis perpendicular to the rod that passes through the left end of the rod.

Accepted Answer

The answer of A long, thin rod of uniform cross...

Question 12

A pulley has an initial angular speed of 12.5 rad/s and a constant angular acceleration of 3.41 rad/s². Through what angle does the pulley turn in 5.26 s?

Accepted Answer

The answer of A pulley has an initial angular speed...

Question 13

A 10.0-kg rod has a varying linear density that is symmetric about the midpoint of the rod. The moment of inertia about an axis perpendicular to the rod that passes through the rod 30.0 cm from its midpoint is 5.00 kg∙m². What is the moment of inertia about an axis perpendicular to the rod that passes through the rod 20.0 cm from its midpoint?

Accepted Answer

The moment of inertia depends on the distance from the axis to the center of mass. Since the density is symmetric, the moment of inertia decreases as the axis moves closer to the center of mass.

Question 14

A two-dimensional object placed in the xy-plane has three forces acting on it: a force of 3.0 N along the x-axis acting at a point (3 m, 4 m); A force of 2.0 N along the y-axis acting at (-2 m, 5 m); and a force of 5 N in the negative x direction acting at (-2 m, -3 m). What is the net torque about the point (-1 m, 1 m)?

Accepted Answer

The net torque is calculated by summing the torques due to each force about the point (-1 m, 1 m). The torque due to each force is given by τ = r × F, where r is the position vector from the point to the point of application of the force, and F is the force vector. Calculating each torque and summing them gives a net torque of 31 N m clockwise.

Question 15

A wheel rotates through an angle of 320° as it slows down from 78.0 rpm to 22.8 rpm. What is the magnitude of the average angular acceleration of the wheel?

Accepted Answer

The answer of A wheel rotates through an angle of...

Question 16

A person pushes on a doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s². What is the moment of inertia of the door about the hinges?

Accepted Answer

The moment of inertia of the door about the hinges is given by:$$I = \frac{F \cdot d}{\alpha}$$where:- F is the force applied to the doorknob (5.00 N)- d is the distance from the doorknob to the axis of the hinges (0.800 m)- α is the angular acceleration of the door (2.00 rad/s^2)Plugging in these values, we get:$$I = \frac{5.00 N \cdot 0.800 m}{2.00 rad/s^2} = 2.00 kg∙m^2$$

Question 17

A man is holding an 8.00-kg vacuum cleaner at arm's length, a distance of 0.550 m from his shoulder. What is the torque on the shoulder joint if the arm is horizontal?

Accepted Answer

The answer of A man is holding an 8.00-kg vacuum...

Question 18

In an effort to loosen the bolt on the wheel of a car, a man with a mass of 70 kg steps on the end of a 50-cm tire iron which is extending horizontally from the bolt. How much torque is he applying to the bolt?

Accepted Answer

The torque is equal to the force applied multiplied by the distance from the point of application to the axis of rotation. In this case, the force is the man's weight, which is 70 kg * 9.8 m/s^2 = 686 N. The distance from the point of application to the axis of rotation is 50 cm = 0.5 m. Therefore, the torque is 686 N * 0.5 m = 343 Nm.

Question 19

A horizontal 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam, a 200-N forces pushes downward. At the west end of the beam, a 200-N force pushed upward. What is the angular acceleration of the beam?

Accepted Answer

The answer of A horizontal 2.00-m long, 5.00-kg uniform beam...

Question 20

A flywheel rotating at 640 rev/min is brought to rest with a uniform deceleration of 2.0 rad/s². How many revolutions does it make before coming to rest?

Accepted Answer

The answer of A flywheel rotating at 640 rev/min is...

Quiz 10: Rotational Motion

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s2. Through what angle has the wheel turned when its angular speed reaches 72.0 rad/s?