PHY 107 - CH11 Rolling, Torque, Angular Momentum Notes

Recap from Last Lecture

• Force analysis for rolling motion

• Static friction force for rotation

• No heat loss allows for the use of mechanical energy conservation.

• The direction of friction opposes slipping.

Newton’s 2nd Law for Rotation

• $\tau_{net} = I \alpha$

• $rF_f = I \alpha$

• The frictional force produces rotation.

• Without friction, the object will slide.

• $a_{com,x}$ is negative, but $\alpha$ is positive.

• $RF<em>f = I</em>{com} (-\frac{a_{com,x}}{R})$

• $a_{com,x} = -R\alpha$

• $F<em>{net, x} = Ma</em>{com, x}$

• $F<em>f = \frac{I</em>{com}}{R^2} \alpha$

Example: Ball Rolling Down a Ramp

• A uniform ball with mass $M = 6 kg$ and radius $R$ rolls smoothly from rest down a ramp at an angle $\theta = 30°$. The ball descends a vertical height $h = 1.2 m$.

  • (a) What is its speed at the bottom?

  • (b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp?

• Given:

  • $M = 6 kg$

  • $\theta = 30°$

  • $h = 1.2 m$

  • $I_{com} = \frac{2}{5}MR^2$ (for a sphere)

• Part (a): Find $v_{com}$.

  • Use kinetic energy converted from gravitational potential energy.

• Part (b): Find $F_f$.

  • Use the equation derived from the previous slide.

Angular Momentum

• $\vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$

• Magnitude: $l = rp \sin{\phi} = rmv \sin{\phi}$

• Angular momentum is defined as: $\vec{l} = \vec{r} \times \vec{p}$ where $\vec{r}$ is the position vector and $\vec{p}$ is the linear momentum.

• Unit: kg·m²/s

Three ways of Calculating Angular Momentum Magnitude

• Method-1: Use the equation $l = r p \sin{\phi}$.

• Method-2: Find the rotation arm perpendicular to the linear momentum.

• Method-3: Find the linear momentum component perpendicular to the rotation arm.

Angular Momentum and Right Hand Rule

• $\vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$

• Angular momentum has meaning only with respect to a specified origin.

• It is always perpendicular to the plane formed by the position and linear momentum vectors.

• Example-1:

  • Angular momentum points into the screen.

Angular Momentum Example

• Problem: Which particle(s) have negative angular momentum (into the screen) if the center dot (O) is the rotation center?

• Answer: Particle-2 and Particle-4

Single Particle Example

• A particle $m = 1kg$ is passing above the horizontal at a distance of $d = 10 m$, with an instantaneous velocity of $\vec{v} = 10 \hat{i} m/s$ along the horizontal direction. Calculate its angular momentum about the rotation origin (O) (magnitude and direction).

Angular Momentum in particle system

• Sum the angular momenta of the particles to find the angular momentum of a system of particles: $\vec{L} = \sum<em>i \vec{l</em>i}$

• Two particle example: $\vec{L} = \vec{l<em>1} + \vec{l</em>2} = \vec{r<em>1} \times \vec{p</em>1} + \vec{r<em>2} \times \vec{p</em>2}$

Angular Momentum - Example

• In the instant of figure below, two particles move in an xy plane. Particle P1 has 12 kgm/s, and it is at distance d1 = 1 m from point O. Particle P2 has a linear momentum of 9 kgm/s, and it is at distance d2 = 3 m from point O.

• What are the (a) magnitude and (b) direction of the net angular momentum of the two particles about O?

• $\vec{l<em>1} = \vec{r</em>1} \times \vec{p<em>1} = d</em>1 p_1(-\hat{k})$

• $\vec{l<em>2} = \vec{r</em>2} \times \vec{p<em>2} = d</em>2 p_2(+\hat{k})$

• $\vec{L} = \vec{l<em>1} + \vec{l</em>2} = d<em>1 p</em>1(-\hat{k}) + d<em>2 p</em>2(+\hat{k}) = (d<em>2 p</em>2 - d<em>1 p</em>1) \hat{k} = [(3m)(9 kg \cdot m/s) - (1.0m)(12 kg \cdot m/s)] \hat{k} = 15 \hat{k} (kg \cdot m^2/s)$

Translational Motion and Angular Momentum

• Translational motion also has angular momentum with respect to a specified origin.

• $\vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$

Angular Velocity Format

• Two particle example:

• In this case, all angles are 90 degrees, we can get the amplitude:

• $\vec{L} = \vec{l<em>1} + \vec{l</em>2}$

• $\vec{L} = \vec{r<em>1} \times m</em>1 \vec{v<em>1} + \vec{r</em>2} \times m<em>2 \vec{v</em>2}$

• $L = m<em>1 d</em>1 v<em>1 + m</em>2 d<em>2 v</em>2$

• $L = m<em>1 d</em>1 (\omega<em>1 d</em>1) + m<em>2 d</em>2 (\omega<em>2 d</em>2)$

• $v = \omega r$

• $L = m<em>1 d</em>1^2 \omega<em>1 + m</em>2 d<em>2^2 \omega</em>2$

Example

• Three particles of mass $m = 23 g$ are fastened to three rods of length $d = 12 cm$ and rod has a mass of $M = 0.5 kg$. The rigid assembly rotates around point O at the angular speed $\omega = 0.85 rad/s$. About O, what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the magnitude of the angular momentum of the assembly?

• $I<em>{total} = I</em>{rod} + I_{particles}$

Rigid Body Rotation

• Torque and angular momentum must be measured relative to the same origin.

• We can find the angular momentum of a rigid body through summation:

• $L = \sum l<em>i = \sum r</em>i p<em>i = \sum r</em>i (m<em>i v</em>i) = \sum r<em>i (m</em>i \omega r<em>i) = \sum m</em>i r<em>i^2 \omega = (\sum m</em>i r_i^2) \omega$

• Since the sum is the rotational inertia I of the body, we can write: $\vec{L} = I \vec{\omega}$

Example: Merry-Go-Round Carrying a Particle

• A merry-go-round (radius of R) is rotating together with a particle with an angular speed of $\omega$. The particle is at the edge of the merry-go-round. What is the system’s total angular momentum?

• Method one: treat them as two individual:

  • Disk: $L<em>1 = I</em>1 \omega = \frac{1}{2} M R^2 \omega$

  • Particle: $L<em>2 = I</em>2 \omega = M R^2 \omega$

  • Total: $L = L<em>1 + L</em>2 = \frac{1}{2} M R^2 \omega + M R^2 \omega = (\frac{1}{2} M R^2 + M R^2) \omega \hat{k}$

• Method two: treat them as whole system

  • $I<em>{total} = I</em>{disk} + I_{particle} = \frac{1}{2} M R^2 + M R^2$

  • $L = I \omega = (\frac{1}{2} M R^2 + M R^2) \omega = (\frac{1}{2} M R^2 + M R^2) \omega \hat{k}$

Force analysis for rolling motion

• Static friction force for rotation

• No heat loss allows for the use of mechanical energy conservation.

• The direction of friction opposes slipping.

Newton’s 2nd Law for Rotation

$\tau_{net} = I \alpha$

$r F_f = I \alpha$

• The frictional force produces rotation.

• Without friction, the object will slide.

• $a_{com,x}$ is negative, but $\alpha$ is positive.

• $R F<em>f = I</em>{com} \left(-\frac{a_{com,x}}{R}\right)$

• $a_{com,x} = -R \alpha$

• $F<em>{net, x} = M a</em>{com, x}$

• $F<em>f = \frac{I</em>{com}}{R^2} \alpha$

Example: Ball Rolling Down a Ramp

A uniform ball with mass $M = 6 \text{ kg}$ and radius $R$ rolls smoothly from rest down a ramp at an angle $\theta = 30^{\circ}$. The ball descends a vertical height $h = 1.2 \text{ m}$.

(a) What is its speed at the bottom?

(b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp?

Given:

• $M = 6 \text{ kg}$

• $\theta = 30^{\circ}$

• $h = 1.2 \text{ m}$

• $I_{com} = \frac{2}{5} M R^2$ (for a sphere)

Part (a): Find $v_{com}$.

• Use kinetic energy converted from gravitational potential energy.

Part (b): Find $F_f$.

• Use the equation derived from the previous slide.

Angular Momentum

$\vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$

• Magnitude: $l = r p \sin{\phi} = r m v \sin{\phi}$

• Angular momentum is defined as: $\vec{l} = \vec{r} \times \vec{p}$ where $\vec{r}$ is the position vector and $\vec{p}$ is the linear momentum.

Unit: kg·m²/s

Three ways of Calculating Angular Momentum Magnitude

1. Use the equation $l = r p \sin{\phi}$.

2. Find the rotation arm perpendicular to the linear momentum.

3. Find the linear momentum component perpendicular to the rotation arm.

Angular Momentum and Right Hand Rule

• $\vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$

• Angular momentum has meaning only with respect to a specified origin.

• It is always perpendicular to the plane formed by the position and linear momentum vectors.

Example-1:

• Angular momentum points into the screen.

Angular Momentum Example

Problem: Which particle(s) have negative angular momentum (into the screen) if the center dot (O) is the rotation center?

Answer: Particle-2 and Particle-4

Single Particle Example

A particle $m = 1 \text{ kg}$ is passing above the horizontal at a distance of $d = 10 \text{ m}$, with an instantaneous velocity of $\vec{v} = 10 \hat{i} \text{ m/s}$ along the horizontal direction. Calculate its angular momentum about the rotation origin (O) (magnitude and direction).

Angular Momentum in particle system

• Sum the angular momenta of the particles to find the angular momentum of a system of particles: $\vec{L} = \sum<em>i \vec{l</em>i}$

• Two particle example: $\vec{L} = \vec{l<em>1} + \vec{l</em>2} = \vec{r<em>1} \times \vec{p</em>1} + \vec{r<em>2} \times \vec{p</em>2}$

Angular Momentum - Example

• In the instant of figure below, two particles move in an xy plane. Particle P1 has 12 kg·m/s, and it is at distance $d<em>1 = 1 \text{ m}$ from point O. Particle P2 has a linear momentum of 9 kg·m/s, and it is at distance $d</em>2 = 3 \text{ m}$ from point O.

What are the (a) magnitude and (b) direction of the net angular momentum of the two particles about O?

• $\vec{l<em>1} = \vec{r</em>1} \times \vec{p<em>1} = d</em>1 p_1 (-\hat{k})$

• $\vec{l<em>2} = \vec{r</em>2} \times \vec{p<em>2} = d</em>2 p_2 (+\hat{k})$

• \vec{L} = \vec{l1} + \vec{l2} = d1 p1 (-\hat{k}) + d2 p2 (+\hat{k}) = (d2 p2 - d1 p1) \hat{k} = [(3 \text{ m})(9 \text{ kg} \cdot \text{ m/s}) - (1.0\text{ m})(12 \text{ kg} \cdot \text{ m/s})] \hat{k} = 15 \hat{k} (\text{kg} \cdot \text{m}^2/s)$

Translational Motion and Angular Momentum

• Translational motion also has angular momentum with respect to a specified origin.

• \vec{l} = \vec{r} \times \vec{p} = (m \vec{r} \times \vec{v})$</p></li></ul><p>Angular Velocity Format</p><ul><li><p>Two particle example:</p><ul><li><p>In this case, all angles are 90 degrees, we can get the amplitude:</p></li></ul></li></ul><p>$\vec{L} = \vec{l1} + \vec{l2}
  \vec{L} = \vec{r1} \times m1 \vec{v1} + \vec{r2} \times m2 \vec{v2}
  L = m1 d1 v1 + m2 d2 v2
  L = m1 d1 (\omega1 d1) + m2 d2 (\omega2 d2)
  v = \omega r
  L = m1 d1^2 \omega1 + m2 d2^2 \omega2$</p><p>Example</p><ul><li><p>Three particles of mass$m = 23 \text{ g}$are fastened to three rods of length$d = 12 \text{ cm}$and rod has a mass of$M = 0.5 \text{ kg}$. The rigid assembly rotates around point O at the angular speed$\omega = 0.85 \text{ rad/s}$. About O, what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the magnitude of the angular momentum of the assembly?</p></li><li><p>$I{total} = I{rod} + I_{particles}

Rigid Body Rotation

• Torque and angular momentum must be measured relative to the same origin.

• We can find the angular momentum of a rigid body through summation:
  L = \sum li = \sum ri pi = \sum ri (mi vi) = \sum ri (mi \omega ri) = \sum mi ri^2 \omega = \left(\sum mi r_i^2\right) \omega$</p></li><li><p>Since the sum is the rotational inertia I of the body, we can write:$\vec{L} = I \vec{\omega}$</p></li></ul><p>Example: Merry-Go-Round Carrying a Particle</p><ul><li><p>A merry-go-round (radius of R) is rotating together with a particle with an angular speed of$\omega$. The particle is at the edge of the merry-go-round. What is the system’s total angular momentum?</p></li><li><p>Method one: treat them as two individual:</p><ul><li><p>Disk:$L1 = I1 \omega = \frac{1}{2} M R^2 \omega$</p></li><li><p>Particle:$L2 = I2 \omega = M R^2 \omega$</p></li><li><p>Total:$L = L1 + L2 = \frac{1}{2} M R^2 \omega + M R^2 \omega = \left(\frac{1}{2} M R^2 + M R^2\right) \omega \hat{k}$</p></li></ul></li><li><p>Method two: treat them as whole system</p><ul><li><p>$I{total} = I{disk} + I_{particle} = \frac{1}{2} M R^2 + M R^2$</p></li><li><p>$L = I \omega = \left(\frac{1}{2} M R^2 + M R^2\right) \omega = \left(\frac{1}{2} M R^2 + M R^2\right) \omega \hat{k}$$