A wheel turns through an angle of 6 rad in 3 s at a constant rate. What is its average angular velocity?
2 rad/s
0.5 rad/s
18 rad/s
9 rad/s
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WhyAverage angular velocity equals change in angle divided by time: 6 rad / 3 s = 2 rad/s.
2Multiple choice · Easy
Which quantity is the rotational analog of force in Newton's second law?
Torque
Moment of inertia
Angular momentum
Angular velocity
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WhyTorque plays the role of force in the rotational form of Newton's second law (net torque = I times angular acceleration).
3Multiple choice · Easy
A force of 10 N is applied perpendicular to a wrench at a distance of 0.30 m from the pivot. What is the magnitude of the torque about the pivot?
3 N m
30 N m
0.03 N m
10 N m
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WhyFor a perpendicular force, torque = r times F = 0.30 m times 10 N = 3 N m.
4Multiple choice · Easy
To produce the largest torque with a given force on a door, where should the force be applied and in what direction?
Far from the hinge, perpendicular to the door
Close to the hinge, perpendicular to the door
Far from the hinge, parallel to the door
Close to the hinge, parallel to the door
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WhyTorque increases with lever arm and is maximized when the force is perpendicular to the radius, so push far from the hinge and perpendicular to the door.
5Multiple choice · Easy
A rigid body is in rotational equilibrium. What must be true about the torques acting on it?
The net torque is zero
The total torque equals the moment of inertia
Only one torque acts on it
The angular velocity must be zero
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WhyRotational equilibrium means the net torque about any axis is zero, so clockwise and counterclockwise torques balance.
6Multiple choice · Easy
A disk speeds up from rest to 10 rad/s in 5 s with constant angular acceleration. What is its angular acceleration?
2 rad/s^2
50 rad/s^2
0.5 rad/s^2
5 rad/s^2
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WhyAngular acceleration = change in angular velocity / time = (10 rad/s - 0) / 5 s = 2 rad/s^2.
7Fill in the blank · Easy
The moment of inertia of a point mass m at distance r from the axis of rotation is I = m r^.
Answer:
2
WhyFor a point mass, I = m r^2; the exponent on the radius is 2.
8Multiple choice · Medium
A solid disk of moment of inertia 0.50 kg m^2 experiences a net torque of 4.0 N m. What is its angular acceleration?
8.0 rad/s^2
2.0 rad/s^2
0.125 rad/s^2
4.5 rad/s^2
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WhyUsing net torque = I times alpha, alpha = 4.0 N m / 0.50 kg m^2 = 8.0 rad/s^2.
9Multiple choice · Medium
A point on the rim of a wheel of radius 0.25 m has a tangential (linear) speed of 4.0 m/s. What is the angular velocity of the wheel?
16 rad/s
1.0 rad/s
0.0625 rad/s
4.0 rad/s
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WhyLinear and rotational speed relate by v = r times omega, so omega = v / r = 4.0 m/s / 0.25 m = 16 rad/s.
10Multiple choice · Medium
A force of 20 N is applied at the end of a 0.40 m rod at an angle of 30 degrees to the rod. What is the magnitude of the torque about the other end?
4.0 N m
8.0 N m
6.9 N m
2.0 N m
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WhyTorque = r times F times sin(theta) = 0.40 m times 20 N times sin(30 deg) = 0.40 times 20 times 0.5 = 4.0 N m.
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