Unit 6: Conservation of Angular Momentum
Rotational Energy, Angular Momentum Systems, and Keplerian Motion
Section 6.1
Rotational Kinetic Energy
For a rigid body rotating with angular velocity \( \omega \), the rotational kinetic energy is \( K_{rot} = \displaystyle\frac{1}{2}I\omega^2 \). For rolling without slipping, \( K_{total} = \displaystyle\frac{1}{2}Mv_{cm}^2 + \displaystyle\frac{1}{2}I_{cm}\omega^2 \).
Q1: A solid sphere (\( I = \displaystyle\frac{2}{5}MR^2 \)) rolls without slipping. What fraction of its total kinetic energy is rotational?
\( K_{trans} = \displaystyle\frac{1}{2}Mv^2 \).
\( K_{rot} = \displaystyle\frac{1}{2}(\displaystyle\frac{2}{5}MR^2)(\displaystyle\frac{v}{R})^2 = \displaystyle\frac{1}{5}Mv^2 \).
\( K_{total} = 0.5Mv^2 + 0.2Mv^2 = 0.7Mv^2 \).
Ratio: \( \displaystyle\frac{0.2}{0.7} = \displaystyle\frac{2}{7} \).
Correct Answer: B
\( K_{rot} = \displaystyle\frac{1}{2}(\displaystyle\frac{2}{5}MR^2)(\displaystyle\frac{v}{R})^2 = \displaystyle\frac{1}{5}Mv^2 \).
\( K_{total} = 0.5Mv^2 + 0.2Mv^2 = 0.7Mv^2 \).
Ratio: \( \displaystyle\frac{0.2}{0.7} = \displaystyle\frac{2}{7} \).
Correct Answer: B
Q2: If the angular velocity of a rotating object is doubled, the rotational kinetic energy increases by a factor of:
Since \( K_{rot} = \displaystyle\frac{1}{2}I\omega^2 \), the energy is proportional to the square of angular velocity. \( 2^2 = 4 \).
Correct Answer: B
Correct Answer: B
Section 6.2
Work and Rotational Power
Work done by a torque is \( W = \int \tau \, d\theta \). The rotational work-energy theorem states \( W_{net} = \Delta K_{rot} \). Power is the rate of rotational work: \( P = \tau\omega \).
Q1: A motor provides \( 1000 \text{ W} \) of power to a shaft rotating at \( 10 \text{ rad/s} \). What is the torque produced?
Using the power formula: \( P = \tau\omega \).
\( 1000 = \tau(10) \Rightarrow \tau = 100 \text{ Nm} \).
Correct Answer: B
\( 1000 = \tau(10) \Rightarrow \tau = 100 \text{ Nm} \).
Correct Answer: B
Q2: A frictional torque of \( 2 \text{ Nm} \) acts on a disk (\( I = 4 \text{ kgm}^2 \)) starting at \( 10 \text{ rad/s} \). How far (in radians) does it rotate before stopping?
Work-Energy: \( -\tau\Delta\theta = 0 – \displaystyle\frac{1}{2}I\omega_i^2 \).
\( -2\Delta\theta = -0.5(4)(10^2) = -200 \).
\( \Delta\theta = 100 \text{ rad} \).
Correct Answer: B
\( -2\Delta\theta = -0.5(4)(10^2) = -200 \).
\( \Delta\theta = 100 \text{ rad} \).
Correct Answer: B
Section 6.3
Angular Momentum
Angular momentum is \( \vec{L} = I\vec{\omega} \) for rigid bodies and \( \vec{L} = \vec{r} \times \vec{p} \) for point particles. In the absence of external torque, \( L \) is conserved.
Q1: An ice skater pulls her arms in, reducing her rotational inertia by half. Her angular velocity will:
By conservation of angular momentum: \( I_1\omega_1 = I_2\omega_2 \).
If \( I_2 = 0.5I_1 \), then \( \omega_2 = 2\omega_1 \).
Correct Answer: C
If \( I_2 = 0.5I_1 \), then \( \omega_2 = 2\omega_1 \).
Correct Answer: C
Q2: A particle of mass \( m \) moves at speed \( v \) along the line \( y = b \). What is its angular momentum about the origin?
Angular momentum \( L = r p \sin\phi \), where \( r \sin\phi \) is the perpendicular distance (lever arm). For a particle moving on \( y=b \), the lever arm to the origin is always \( b \). Thus, \( L = mvb \).
Correct Answer: B
Correct Answer: B
Rotational Energy & Momentum Recap
Rotational KE\( K_{rot} = \displaystyle\frac{1}{2}I\omega^2 \)
Rotational Work\( W = \int \tau \, d\theta \)
Rotational Power\( P = \tau\omega \)
Angular Momentum\( \vec{L} = I\vec{\omega} \)
Point Particle L\( \vec{L} = \vec{r} \times \vec{p} \)
Angular Impulse\( \Delta \vec{L} = \int \vec{\tau} dt \)
Unit 6: Rotational Motion (Part 2)
Conservation of Angular Momentum, Rolling, and Satellite Motion
Section 6.4
Conservation of Angular Momentum
If the net external torque is zero (\( \sum \tau_{ext} = 0 \)), the total angular momentum of the system is conserved: \( L_i = L_f \). Reference file: “截圖 2026-05-12 晚上7.41.43.png”.
Q1: A figure skater rotates with an angular velocity \( \omega_0 \). When she pulls her arms in, her rotational inertia decreases to \( \displaystyle\frac{1}{3} \) of its original value. What is her new angular velocity?
Conservation of angular momentum: \( I_i\omega_i = I_f\omega_f \).
\( I_0\omega_0 = (\displaystyle\frac{1}{3}I_0)\omega_f \).
Solving for \( \omega_f \): \( \omega_f = 3\omega_0 \).
Correct Answer: C
\( I_0\omega_0 = (\displaystyle\frac{1}{3}I_0)\omega_f \).
Solving for \( \omega_f \): \( \omega_f = 3\omega_0 \).
Correct Answer: C
Q2: A small mass \( m \) moving with velocity \( v \) hits a stationary rod length \( L \) (pivoted at the top) and sticks to the end. Which quantity is conserved during the collision?
The pivot exerts an external force, so linear momentum is not conserved. The collision is inelastic, so kinetic energy is lost. However, the pivot exerts no torque about itself, so angular momentum about the pivot is conserved.
Correct Answer: C
Correct Answer: C
Q3: A planet moves in an elliptical orbit. As it moves from aphelion (far) to perihelion (near), its angular momentum and speed behave as follows:
Gravity acts as a central force toward the star, providing zero torque. Thus, \( L \) is conserved. Since \( L = mvr \sin\phi \), as radius \( r \) decreases, the speed \( v \) must increase.
Correct Answer: B
Correct Answer: B
Section 6.5
Rolling Motion
Rolling without slipping combines translation and rotation: \( v_{cm} = R\omega \). The total kinetic energy is \( K_{total} = \displaystyle\frac{1}{2}Mv_{cm}^2 + \displaystyle\frac{1}{2}I_{cm}\omega^2 \).
Q1: A solid cylinder of mass \( M \) and radius \( R \) rolls without slipping. What fraction of its total kinetic energy is translational? (\( I = \displaystyle\frac{1}{2}MR^2 \))
\( K_{trans} = \displaystyle\frac{1}{2}Mv^2 \).
\( K_{rot} = \displaystyle\frac{1}{2}I\omega^2 = \displaystyle\frac{1}{2}(\displaystyle\frac{1}{2}MR^2)(\displaystyle\frac{v}{R})^2 = \displaystyle\frac{1}{4}Mv^2 \).
\( K_{total} = \displaystyle\frac{1}{2}Mv^2 + \displaystyle\frac{1}{4}Mv^2 = \displaystyle\frac{3}{4}Mv^2 \).
Fraction translational: \( \displaystyle\frac{1/2}{3/4} = \displaystyle\frac{2}{3} \).
Correct Answer: C
\( K_{rot} = \displaystyle\frac{1}{2}I\omega^2 = \displaystyle\frac{1}{2}(\displaystyle\frac{1}{2}MR^2)(\displaystyle\frac{v}{R})^2 = \displaystyle\frac{1}{4}Mv^2 \).
\( K_{total} = \displaystyle\frac{1}{2}Mv^2 + \displaystyle\frac{1}{4}Mv^2 = \displaystyle\frac{3}{4}Mv^2 \).
Fraction translational: \( \displaystyle\frac{1/2}{3/4} = \displaystyle\frac{2}{3} \).
Correct Answer: C
Q2: A ball rolls without slipping on a horizontal surface with speed \( v \). If it rolls up an incline, to what max height \( h \) does it rise? (\( I = \displaystyle\frac{2}{5}MR^2 \))
\( K_{total} = \displaystyle\frac{7}{10}Mv^2 \) (Sum of trans and rot KE).
Energy conservation: \( \displaystyle\frac{7}{10}Mv^2 = Mgh \).
\( h = \displaystyle\frac{7v^2}{10g} \).
Correct Answer: C
Energy conservation: \( \displaystyle\frac{7}{10}Mv^2 = Mgh \).
\( h = \displaystyle\frac{7v^2}{10g} \).
Correct Answer: C
Section 6.6
Orbiting Satellites
Satellites in circular orbits are governed by \( \displaystyle\frac{GMm}{r^2} = \displaystyle\frac{mv^2}{r} \). This leads to an orbital speed \( v = \sqrt{\displaystyle\frac{GM}{r}} \).
Q1: A satellite’s orbital radius is doubled. How does its orbital speed change?
\( v = \sqrt{\displaystyle\frac{GM}{r}} \). If \( r \rightarrow 2r \), then \( v \rightarrow \displaystyle\frac{1}{\sqrt{2}}v \).
Correct Answer: C
Correct Answer: C
Q2: For a satellite in a circular orbit, what is the ratio of potential energy \( U \) to kinetic energy \( K \)?
\( U = -\displaystyle\frac{GMm}{r} \) and \( K = \displaystyle\frac{GMm}{2r} \).
Therefore, \( \displaystyle\frac{U}{K} = -2 \).
Correct Answer: D
Therefore, \( \displaystyle\frac{U}{K} = -2 \).
Correct Answer: D
Rotational Conservation & Orbits Recap
Angular Momentum\( L = I\omega \text{ or } \vec{r} \times \vec{p} \)
Rolling KE\( K = \displaystyle\frac{1}{2}Mv^2 + \displaystyle\frac{1}{2}I\omega^2 \)
No-Slip Link\( v_{cm} = R\omega \)
Orbital Speed\( v = \sqrt{\displaystyle\frac{GM}{r}} \)
Kepler’s 3rd\( T^2 \propto r^3 \)
Satellite Energy\( E = -\displaystyle\frac{GMm}{2r} \)
