Rotational Dynamics

(b) Torque

Torque is the turning effect of a force. For a rigid body $\tau = I\alpha$, so a net torque produces angular acceleration, just as a net force produces linear acceleration in $F = ma$.

(The principle behind all three parts is really the conservation of angular momentum, $L = I\omega = \text{constant}$, which is why $\omega \propto 1/I$.)

(a) Mass at the rim of a flywheel

Moment of inertia is $I = \sum m r^2$. Putting most of the mass at the rim (largest $r$) gives the largest moment of inertia for the same mass. A large $I$ means:

• the wheel strongly opposes any change in its angular speed, so it keeps the engine shaft turning smoothly even though the driving torque comes in jerks;
• it stores more rotational kinetic energy, $\tfrac12 I\omega^2$.

(b) Earth moves faster when it is nearer the Sun

The Sun's gravitational pull on the Earth acts along the line joining them (a central force). Its moment arm about the Sun is zero, so the torque on the Earth about the Sun is zero. Hence the Earth's angular momentum about the Sun stays constant:

$$L = I\omega = mr^2\omega = \text{constant}$$

When the Earth comes closer to the Sun, $r$ decreases, so $I = mr^2$ decreases. To keep $L$ constant, $\omega$ must increase. (This is also Kepler's second law: equal areas in equal times.)

(c) If the Earth shrank suddenly

No external torque acts on the Earth, so its spin angular momentum is conserved:

$$I_1\omega_1 = I_2\omega_2, \qquad I = \tfrac25 MR^2$$

If the Earth shrinks, its mass $M$ stays the same but $R$ decreases, so $I$ decreases and $\omega$ increases. Since the length of the day is $T = \dfrac{2\pi}{\omega}$, the day becomes shorter.

For example, if the radius became half, $I$ would become $\tfrac14$ of its value, $\omega$ would become 4 times, and the day would be $24/4 = 6$ hours.

(a) Rotational kinetic energy

With $L = I\omega$, the rotational kinetic energy is

$$K = \tfrac12 I\omega^2 = \frac{(I\omega)^2}{2I} = \frac{L^2}{2I}$$

(Here $L$ is the angular momentum.)

(a) Moment of inertia

The moment of inertia of a body about an axis is its rotational inertia, i.e. its tendency to oppose a change in its state of rotation about that axis. It equals the sum of the products of the mass of each particle and the square of its distance from the axis:

$$I = \sum m_i r_i^2$$

SI unit: kg m$^2$.

(b) Principle of conservation of angular momentum

If no external torque acts on a system, its total angular momentum remains constant:

$$\tau_{\text{ext}} = 0 \;\Rightarrow\; L = I\omega = \text{constant}, \quad I_1\omega_1 = I_2\omega_2$$

(c) Numerical

Given:

• Moment of inertia of disc, $I_1 = 5\times10^{-4}$ kg m$^2$
• Initial speed, $n_1 = 40$ rpm
• Mass of wax, $m = 0.02$ kg, at distance $r = 0.08$ m from the axis
• New speed, $n_2 = ?$

Moment of inertia of the wax (treated as a point mass):

$$I_{\text{wax}} = mr^2 = 0.02 \times (0.08)^2 = 1.28\times10^{-4}\ \text{kg m}^2$$

New moment of inertia:

$$I_2 = I_1 + I_{\text{wax}} = (5 + 1.28)\times10^{-4} = 6.28\times10^{-4}\ \text{kg m}^2$$

The wax is dropped gently, so there is no external torque and angular momentum is conserved. Since $\omega = 2\pi n$, we can work directly in rpm:

$$I_1 n_1 = I_2 n_2 \;\Rightarrow\; n_2 = \frac{I_1 n_1}{I_2} = \frac{5\times10^{-4}\times 40}{6.28\times10^{-4}}$$ $$\boxed{n_2 \approx 31.8\ \text{rpm}}$$

(a) Torque

In linear motion force causes linear acceleration ($F = ma$). In rotational motion torque causes angular acceleration ($\tau = I\alpha$). So torque is the rotational analogue of force, and moment of inertia is the analogue of mass.

a) Angular momentum

The angular momentum of a particle about an axis is the moment of its linear momentum about that axis: $\vec L = \vec r \times \vec p$. For a rigid body rotating about a fixed axis,

$$L = I\omega$$

SI unit: kg m$^2$ s$^{-1}$ (equivalently J s).

b) Numerical

Given:

• Initial speed, $n_1 = 900$ rev/min $= 15$ rev/s, so $\omega_1 = 2\pi \times 15 = 30\pi$ rad s$^{-1}$ $\approx 94.2$ rad s$^{-1}$
• Final speed, $n_2 = 600$ rev/min $= 10$ rev/s, so $\omega_2 = 2\pi \times 10 = 20\pi$ rad s$^{-1}$ $\approx 62.8$ rad s$^{-1}$
• Time, $t = 10$ s

i) Angular acceleration

$$\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{20\pi - 30\pi}{10} = -\pi\ \text{rad s}^{-2}$$ $$\boxed{\alpha \approx -3.14\ \text{rad s}^{-2}}$$

The negative sign shows that this is a retardation of 3.14 rad s$^{-2}$.

ii) Number of revolutions in 10 s

$$\theta = \left(\frac{\omega_1 + \omega_2}{2}\right)t = \frac{30\pi + 20\pi}{2}\times 10 = 250\pi\ \text{rad}$$ $$N = \frac{\theta}{2\pi} = \frac{250\pi}{2\pi} = \boxed{125\ \text{revolutions}}$$

iii) Additional time to come to rest

Now $\omega_1' = 20\pi$ rad s$^{-1}$, $\omega_2' = 0$, $\alpha = -\pi$ rad s$^{-2}$:

$$t' = \frac{\omega_2' - \omega_1'}{\alpha} = \frac{0 - 20\pi}{-\pi} = \boxed{20\ \text{s}}$$

(B) $\sqrt{2EI}$

$E = \tfrac12 I\omega^2 = \dfrac{L^2}{2I}$, so $L^2 = 2EI$ and $L = \sqrt{2EI}$.

a) Moment of inertia

The moment of inertia of a body about an axis is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis, $I = \sum m_i r_i^2$. It measures the body's opposition to a change in its rotational motion. SI unit: kg m$^2$.

b) Moment of inertia of a uniform rod about an axis through one end

      A
      |
      |<------------------ L ------------------>|
      |==========================[dx]===========|
      |<-------- x -------->|
      |
      B
   axis AB (perpendicular to the rod, through end)

1. Let the rod have mass $M$ (written $m$ in the question) and length $L$. Its mass per unit length is $\lambda = \dfrac{M}{L}$.
2. Take a small element of length $dx$ at distance $x$ from the axis AB. Its mass is

$$dm = \frac{M}{L}\,dx$$

3. Moment of inertia of this element about AB:

$$dI = dm\,x^2 = \frac{M}{L}x^2\,dx$$

4. The whole rod extends from $x = 0$ to $x = L$, so

$$I = \int_0^L \frac{M}{L}x^2\,dx = \frac{M}{L}\left[\frac{x^3}{3}\right]_0^L = \frac{M}{L}\cdot\frac{L^3}{3}$$

5. Hence

$$\boxed{I_{AB} = \frac{ML^2}{3}}$$

c) Numerical

Given: $I = 0.32$ kg m$^2$, $\omega = 120$ rad s$^{-1}$, power $P = 50$ W.

i) Kinetic energy of the flywheel

$$K = \tfrac12 I\omega^2 = \tfrac12 \times 0.32 \times (120)^2 = 0.16 \times 14400$$ $$\boxed{K = 2304\ \text{J} \approx 2.30\times10^3\ \text{J}}$$

ii) Frictional couple

The wheel turns steadily ($\alpha = 0$), so the motor's torque just balances the frictional couple. All the motor power is used against friction:

$$P = \tau\omega \;\Rightarrow\; \tau = \frac{P}{\omega} = \frac{50}{120}$$ $$\boxed{\tau \approx 0.417\ \text{N m}}$$

(B) $\dfrac{ML^2}{4}$

About an end: $I_{\text{end}} = \dfrac{ML^2}{3}$. About the centre: $I_{\text{c}} = \dfrac{ML^2}{12}$.

$$\frac{ML^2}{3} - \frac{ML^2}{12} = \frac{4ML^2 - ML^2}{12} = \frac{ML^2}{4}$$

(This also follows from the parallel axis theorem: $M(L/2)^2 = ML^2/4$.)

a) Ballet dancer stretching and folding her arms

While spinning on her toe, no external torque acts on the dancer about the vertical axis (friction at a point on the axis gives negligible torque). So her angular momentum is conserved:

$$L = I\omega = \text{constant}$$

• When she stretches her arms, mass moves away from the axis, $I$ increases, so $\omega$ decreases (she spins slower).
• When she folds her arms, $I$ decreases, so $\omega$ increases (she spins faster).

In this way she controls her speed of spin without any push from outside.

b) Relation between torque and moment of inertia

1. Consider a rigid body rotating about a fixed axis with angular acceleration $\alpha$. It is made of particles of masses $m_1, m_2, \ldots$ at distances $r_1, r_2, \ldots$ from the axis.
2. The linear (tangential) acceleration of the $i$-th particle is $a_i = r_i\alpha$.
3. The force needed for this acceleration is $F_i = m_i a_i = m_i r_i \alpha$.
4. The torque of this force about the axis is

$$\tau_i = F_i r_i = m_i r_i^2 \alpha$$

5. All particles have the same $\alpha$, so the total torque is

$$\tau = \sum m_i r_i^2\,\alpha = \left(\sum m_i r_i^2\right)\alpha$$

6. Since $\sum m_i r_i^2 = I$,

$$\boxed{\tau = I\alpha}$$

This is the rotational form of Newton's second law.

c) Why a long wrench is preferred

Torque is $\tau = F \times r$, where $r$ is the arm length. For the torque needed to turn a nut, a longer arm needs a smaller force, so a long wrench makes it easier to open a tight nut.

(b) Axis of rotation

From $I = Mk^2$, $k = \sqrt{I/M}$. The radius of gyration depends on how the mass is distributed about the axis, so it changes with the position and direction of the axis of rotation. It does not depend on angular speed, angular acceleration or $g$.

a) Earth moves faster when nearer the Sun

The gravitational force of the Sun on the Earth acts along the line joining them, so it has zero torque about the Sun. Therefore the Earth's angular momentum about the Sun is conserved:

$$L = I\omega = mr^2\omega = \text{constant}$$

When the Earth comes closer to the Sun, $r$ decreases, so $I = mr^2$ decreases. For $L$ to stay constant, the angular velocity $\omega$ must increase.

b) Numerical

Given: $I = 10$ kg m$^2$, $\omega = 20$ rad s$^{-1}$.

i) Rotational kinetic energy

$$K = \tfrac12 I\omega^2 = \tfrac12 \times 10 \times (20)^2 = \boxed{2000\ \text{J}}$$

ii) Revolutions per second

$$\omega = 2\pi f \;\Rightarrow\; f = \frac{\omega}{2\pi} = \frac{20}{2\times3.14}$$ $$\boxed{f \approx 3.18\ \text{rev s}^{-1}}$$

(c) Angular momentum

Since $\tau = \dfrac{dL}{dt}$, zero net torque gives $\dfrac{dL}{dt}=0$, so the angular momentum $L = I\omega$ stays constant. Force, linear momentum and angular acceleration need not be constant (angular acceleration is actually zero, but it is not the conserved quantity).

(a) Radius of gyration

The radius of gyration of a body about an axis is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia as the body has about that axis: $I = Mk^2$.

(b) Moment of inertia of a thin rod about its centre

Let the rod have mass $M$ and length $l$. Take an element of length $dx$ at distance $x$ from the centre, of mass $dm = \dfrac{M}{l}dx$. Then

$$I = \int_{-l/2}^{l/2} x^2\,\frac{M}{l}\,dx = \frac{M}{l}\cdot\frac{2}{3}\left(\frac{l}{2}\right)^3 = \frac{Ml^2}{12}$$

(c) Ratio of kinetic energies

There is no external torque on the teacher and platform, so angular momentum is conserved:

$$I_1\omega_1 = I_2\omega_2$$

Kinetic energy is $K = \tfrac12 I\omega^2 = \tfrac12 L\omega$ with $L$ constant, so

$$\frac{K_2}{K_1} = \frac{\omega_2}{\omega_1} = \frac{5.0}{1.5} = 3.33$$

The final kinetic energy is about $3.3$ times the initial value (the extra energy comes from the work done by the teacher pulling his arms in).

(c) zero

The centre of mass is the point about which the weights of all particles balance. Taking moments of $m_i g$ about the centre of mass gives $\sum m_i g\,\vec r_i = g\sum m_i\vec r_i = 0$, because $\sum m_i\vec r_i = 0$ about the centre of mass. So the net gravitational torque about it is zero.

(a) Moment of inertia is the property of a body by which it opposes any change in its rotational motion. For a body, $I=\sum m r^2$ about the axis of rotation, with SI unit $\text{kg m}^2$.

(b) Statement: If no external torque acts on a system, its total angular momentum remains constant.

Explanation: Torque is $\tau = dL/dt$. When $\tau = 0$, $dL/dt=0$, so

$L = I\omega = \text{constant} \;\Rightarrow\; I_1\omega_1 = I_2\omega_2$

If $I$ decreases, $\omega$ must increase, and vice versa. For example, a skater spinning with arms stretched out spins faster when she pulls her arms in, because $I$ decreases.

(c) The angular speed is

$\omega = \frac{2\pi\times 4800}{60} = 160\pi = 502.7\ \text{rad/s}$

$P = \tau\omega = 4.30\times 502.7 \approx 2.16\times10^{3}\ \text{W}$

The power output is about 2.16 kW.

(a) Moment of inertia is the property of a body by which it resists any change in its state of rotation about an axis. It is $I = \sum mr^2$ and its SI unit is $\text{kg m}^2$.

(b) Consider a rigid body rotating with angular velocity $\omega$ about a fixed axis. A particle of mass $m_i$ at distance $r_i$ from the axis has speed $v_i = \omega r_i$, so its kinetic energy is

$K_i = \tfrac12 m_iv_i^2 = \tfrac12 m_ir_i^2\omega^2$

Summing over all particles:

$K = \tfrac12\left(\sum m_ir_i^2\right)\omega^2 = \tfrac12 I\omega^2$

(c) With angular momentum $L = I\omega$, the kinetic energy can be written as

$K = \frac12 I\omega^2 = \frac{L^2}{2I}$

For equal $L$, $K\propto 1/I$. Since $I_a>I_b$, body B (with the smaller moment of inertia) has the greater rotational kinetic energy.

(D) angular momentum

For a rigid body rotating about a fixed axis, $L = I\omega$. This is the rotational analogue of linear momentum $p = mv$.

(a) $k^2 = I/m$

The radius of gyration $k$ is the distance from the axis at which the whole mass could be placed to give the same moment of inertia: $I = mk^2$, so $k^2 = I/m$, or $k = \sqrt{I/m}$.

(i) Moment of inertia

The moment of inertia of a body about an axis is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis:

$$I = \sum m_i r_i^2$$

It measures the body's opposition to a change in its rotational motion. SI unit: kg m$^2$.

(ii) Earth moves faster when nearer the Sun

The Sun's gravitational pull acts along the line joining the Sun and the Earth, so its torque about the Sun is zero. Hence the Earth's angular momentum about the Sun is constant:

$$L = I\omega = mr^2\omega = \text{constant}$$

As the Earth comes closer, $r$ decreases, so $I = mr^2$ decreases, and $\omega$ increases to keep $L$ constant.

(iii) If the Earth shrank suddenly

No external torque acts on the Earth, so its spin angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$, with $I = \tfrac25 MR^2$.

On shrinking, $M$ is unchanged but $R$ decreases, so $I$ decreases and $\omega$ increases. Since the length of the day is $T = 2\pi/\omega$, the day becomes shorter.

For example, if the radius became half, $I$ would fall to $\tfrac14$, $\omega$ would become 4 times, and a day would last only $24/4 = 6$ hours.