Rotational Motion and Torque — Hard Practice Quiz
A Physics cheat sheet for Rotational Motion and Torque — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Angular Displacement: Angle \(\theta\) (in radians) is arc length \(s\) divided by radius \(r\).
Average Angular Velocity: Change in angular displacement \(\Delta\theta\) over time \(\Delta t\).
Average Angular Acceleration: Change in angular velocity \(\Delta\omega\) over time \(\Delta t\).
Rotational Kinematics (Velocity-Time): Final angular velocity \(\omega\) given initial \(\omega_0\), acceleration \(\alpha\), and time \(t\).
Rotational Kinematics (Displacement-Time): Angular displacement \(\theta\) given initial velocity, acceleration, and time.
Rotational Kinematics (Velocity-Displacement): Relates velocities, acceleration, and displacement without time.
Tangential Velocity: Linear speed \(v_T\) of a point at radius \(r\) with angular velocity \(\omega\).
Tangential Acceleration: Linear acceleration \(a_T\) of a point at radius \(r\) with angular acceleration \(\alpha\).
Centripetal Acceleration: Radial acceleration component toward the center.
Rolling Motion (No Slipping): Linear velocity and acceleration of the center of mass related to angular quantities.
Torque: The magnitude of torque \(\tau\) is the distance \(r\) from the pivot to the force \(F\), multiplied by the perpendicular component of the force.
Rotational Analog of Newton's Second Law: Net torque \(\tau_{net}\) equals moment of inertia \(I\) multiplied by angular acceleration \(\alpha\).
Moment of Inertia (Point Masses): The moment of inertia \(I\) is the sum of each mass \(m\) times its distance \(r\) from the axis squared.
Angular Velocity (Constant Angular Acceleration): Final angular velocity \(\omega_f\) equals initial \(\omega_i\) plus angular acceleration \(\alpha\) times time \(t\).
Rotational Kinetic Energy: The energy of an object due to its rotation.
Angular Momentum: Angular momentum \(L\) is the moment of inertia \(I\) multiplied by the angular velocity \(\omega\).
Net Torque and Angular Momentum: Net external torque equals the rate of change of angular momentum.
Equilibrium Conditions: For a rigid body to be in equilibrium, the net force and net torque must both be zero.
Center of Gravity: The point where the total weight of the body can be considered to act.
Rotational Work: Work done by a constant torque \(\tau\) rotating an object through angle \(\theta\).
Total Mechanical Energy (Rolling): Sum of translational KE, rotational KE, and gravitational PE.
Conservation of Angular Momentum: If net external torque is zero, total angular momentum is conserved.
Practice quiz
A wheel starts from rest and accelerates with constant angular acceleration $\alpha$. At time $t$, a point on its rim has tangential acceleration $a_T$ and centripetal acceleration $a_c$. If the angular acceleration is doubled to $2\alpha$, what is the ratio of the new centripetal acceleration to the original centripetal acceleration at the same time $t$?
- A) $2$
- B) $4$
- C) $1/2$
- D) $1/4$
Answer: B) $4$
A uniform solid disk of mass $M$ and radius $R$ is rotating with angular velocity $\omega_0$ about an axis through its center. A second identical disk, initially at rest, is dropped onto the first disk. The two disks eventually rotate together with a common final angular velocity $\omega_f$. What fraction of the initial rotational kinetic energy is lost during this process?
- A) $1/4$
- B) $1/2$
- C) $2/3$
- D) $3/4$
Answer: B) $1/2$
A uniform rod of length $L$ and mass $M$ is pivoted at one end and released from rest in a horizontal position. What is the initial angular acceleration of the rod? (Moment of inertia of a rod about one end is $I = \frac{1}{3}ML^2$).
- A) $\frac{3g}{2L}$
- B) $\frac{g}{L}$
- C) $\frac{2g}{3L}$
- D) $\frac{g}{2L}$
Answer: A) $\frac{3g}{2L}$
A solid sphere of mass $M$ and radius $R$ rolls without slipping down an incline of height $h$. What is its speed at the bottom of the incline? (Moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$).
- A) $\sqrt{\frac{2}{5}gh}$
- B) $\sqrt{\frac{10}{7}gh}$
- C) $\sqrt{\frac{7}{10}gh}$
- D) $\sqrt{2gh}$
Answer: B) $\sqrt{\frac{10}{7}gh}$
A uniform ladder of length $L$ and mass $M$ rests against a frictionless wall at an angle $\theta$ with the horizontal ground. The coefficient of static friction between the ladder and the ground is $\mu_s$. What is the minimum angle $\theta$ for the ladder to remain in equilibrium?
- A) $\arctan(\frac{1}{\mu_s})$
- B) $\arctan(\frac{1}{2\mu_s})$
- C) $\arctan(2\mu_s)$
- D) $\arctan(\mu_s)$
Answer: B) $\arctan(\frac{1}{2\mu_s})$
A figure skater is spinning with angular velocity $\omega_0$ and moment of inertia $I_0$. She pulls her arms in, reducing her moment of inertia to $I_f = I_0/3$. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?
- A) $1/3$
- B) $1$
- C) $3$
- D) $9$
Answer: C) $3$
A merry-go-round starts from rest and accelerates with a constant angular acceleration of $\alpha = 0.05 \text{ rad/s}^2$. A child is sitting at a distance $r = 2 \text{ m}$ from the center. How many revolutions has the merry-go-round completed when the child's tangential speed reaches $v_T = 1.5 \text{ m/s}$?
- A) $0.447$ revolutions
- B) $0.895$ revolutions
- C) $1.79$ revolutions
- D) $2.68$ revolutions
Answer: B) $0.895$ revolutions
A uniform rod of mass $M$ and length $L$ is free to rotate about a pivot at its center. A force $F$ is applied perpendicularly to one end of the rod. If the rod's moment of inertia about its center is $I = \frac{1}{12}ML^2$, derive an expression for the angular acceleration $\alpha$ in terms of $F$, $M$, and $L$.
- A) $\frac{6F}{ML}$
- B) $\frac{12F}{ML}$
- C) $\frac{2F}{ML}$
- D) $\frac{F}{ML}$
Answer: A) $\frac{6F}{ML}$
A uniform plank of length $L$ and mass $M$ is supported at two points. One support is at the left end, and the other is at a distance $x$ from the left end. A person of mass $m$ stands at the right end of the plank. If the plank is just about to tip, what is the distance $x$ of the second support from the left end?
- A) $\frac{L(M/2 + m)}{M+m}$
- B) $\frac{L(M+m)}{M/2 + m}$
- C) $\frac{L(M/2)}{M+m}$
- D) $\frac{L(m)}{M+m}$
Answer: A) $\frac{L(M/2 + m)}{M+m}$
A flywheel with a moment of inertia $I = 2.0 \text{ kg} \cdot \text{m}^2$ is initially rotating at $\omega_0 = 10 \text{ rad/s}$. A constant braking torque of $\tau = 5.0 \text{ N} \cdot \text{m}$ is applied. How many revolutions does the flywheel make before coming to rest?
- A) $1.59$ revolutions
- B) $3.18$ revolutions
- C) $6.37$ revolutions
- D) $10.0$ revolutions
Answer: B) $3.18$ revolutions
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