10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Torque is the rotational effectiveness of a force — the turning effect it produces about an axis, given by $\tau = rF\sin\theta$.
For a particle of mass $m$ at distance $r$ from the axis, tangential force $F_t = ma_t = mr\alpha$ (since $a_t = r\alpha$).
Torque due to this force: $\tau = rF_t = r(mr\alpha) = mr^2\alpha$. Summing over all particles of a rigid body: $\tau = \left(\sum mr^2\right)\alpha = I\alpha$
Parallel axis theorem: The moment of inertia about any axis parallel to an axis through the centre of mass equals the moment of inertia about the centre-of-mass axis plus $Md^2$, where $d$ is the perpendicular distance between the two axes: $I = I_{cm} + Md^2$.
Perpendicular axis theorem (applicable to planar/laminar bodies): The moment of inertia about an axis perpendicular to the plane of the body equals the sum of moments of inertia about two mutually perpendicular axes lying in the plane, all three axes intersecting at a common point: $I_z = I_x + I_y$.
Angular momentum is the rotational analogue of linear momentum, defined as $\vec{L} = I\vec{\omega}$ for a rigid body (or $\vec{L} = \vec{r}\times\vec{p}$ for a particle).
Law of conservation: In the absence of any external torque, the total angular momentum of a system remains constant: if $\tau_{ext} = 0$, then $I_1\omega_1 = I_2\omega_2$ — even if the moment of inertia changes (e.g., by redistributing mass), angular velocity adjusts to keep $L$ constant.
The centre of mass is the point at which the entire mass of a system may be considered to be concentrated for describing its overall translational motion — for a system of particles, $\vec{R}_{cm} = \dfrac{\sum m_i\vec{r}_i}{\sum m_i}$.
Internal forces between particles of a system occur in equal-and-opposite (Newton's third law) pairs, so they cancel when summed over the whole system. Only the net external force determines the motion of the centre of mass: $F_{ext} = M a_{cm}$ — this is why the centre of mass moves exactly as a single particle of mass $M$ would, under the net external force alone.
A rolling body simultaneously translates (centre of mass moving at $v$) and rotates (about the centre of mass, at $\omega$), with the rolling condition $v = r\omega$.
Total KE = Translational KE + Rotational KE = $\dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$
Writing $I = mk^2$ (where $k$ is the radius of gyration) and $\omega = v/r$:
$KE = \dfrac{1}{2}mv^2 + \dfrac{1}{2}mk^2\dfrac{v^2}{r^2} = \dfrac{1}{2}mv^2\left(1 + \dfrac{k^2}{r^2}\right)$
A rigid body is in equilibrium under coplanar forces when two independent conditions are simultaneously satisfied:
Translational equilibrium: The vector sum of all forces is zero — $\Sigma \vec{F} = 0$ (no net linear acceleration).
Rotational equilibrium: The vector sum of all torques about any point is zero — $\Sigma \vec{\tau} = 0$ (no net angular acceleration).
The gravitational force between the Sun and a planet acts along the line joining them (a central force), producing zero torque about the Sun (since $\tau = rF\sin\theta$, and the force is radial, $\theta=0$).
Since torque about the Sun is always zero, the planet's angular momentum $L = mvr\sin\phi$ about the Sun remains constant throughout its orbit. When the planet is closer to the Sun (smaller $r$), its speed $v$ must increase to keep $L$ constant — this explains why planets move faster near perihelion (closest approach) than near aphelion (farthest point).
The radius of gyration $k$ of a body about an axis is the distance from that axis at which the entire mass of the body could be concentrated to give the same moment of inertia as the actual body.
It is related to moment of inertia by: $I = Mk^2$, or equivalently, $k = \sqrt{I/M}$.
It provides a convenient way to compare the rotational inertia of bodies of different shapes but equal mass, purely in terms of an equivalent distance from the axis.
For a body rolling down an incline, acceleration is given by $a = \dfrac{g\sin\theta}{1+\frac{k^2}{r^2}}$, where $k$ is the radius of gyration.
A hollow cylinder has its mass concentrated farther from the axis (larger $k^2/r^2 = 1$), while a solid cylinder has mass distributed closer to the axis on average (smaller $k^2/r^2 = 1/2$).
Since a smaller $k^2/r^2$ gives a larger denominator's inverse (larger acceleration), the solid cylinder — with its smaller moment of inertia — accelerates faster down the incline than the hollow cylinder.
Translational equilibrium requires the net force on a body to be zero — the body's centre of mass does not accelerate.
Rotational equilibrium requires the net torque on a body to be zero — the body does not angularly accelerate.
Example: Two equal and opposite forces applied at different points on a rigid rod (a "couple") produce zero net force (translational equilibrium is satisfied — the centre of mass doesn't accelerate), but they produce a non-zero net torque, causing the rod to angularly accelerate (rotational equilibrium is NOT satisfied).
10 questions · 5 marks each · Detailed step-by-step solutions
Angular deceleration: using $\omega = \omega_0 + \alpha t$, with final $\omega=0$: $0 = 30 + \alpha(5) \Rightarrow \alpha = -6\text{ rad/s}^2$
Required torque: $\tau = I\alpha = 2 \times (-6) = -12\text{ N·m}$
A torque of magnitude 12 N·m (opposing the rotation) is required.
Moment of inertia: $I = \frac{1}{2}MR^2 = \frac{1}{2}(4)(0.5)^2 = \frac{1}{2}(4)(0.25) = 0.5\text{ kg·m}^2$
Angular velocity: $\omega = 2\pi n$, where $n = 300/60 = 5$ rev/s. $\omega = 2\pi(5) = 10\pi \approx 31.4\text{ rad/s}$
Angular momentum: $L = I\omega = 0.5 \times 31.4 \approx \mathbf{15.7\text{ kg·m}^2/\text{s}}$
Using energy conservation: loss in PE = gain in total KE (translational + rotational)
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
$v^2 = \dfrac{10gh}{7} = \dfrac{10\times10\times1.4}{7} = \dfrac{140}{7} = 20$
$v = \sqrt{20} \approx \mathbf{4.47\text{ m/s}}$
By conservation of angular momentum (no external torque): $I_1\omega_1 = I_2\omega_2$
$(4)(2) = (1)\omega_2$
$8 = \omega_2$
Her new angular speed is 8 rad/s — a fourfold increase, since her moment of inertia decreased to one-quarter of its original value.
$x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1+m_2} = \dfrac{(3)(0) + (5)(8)}{3+5} = \dfrac{0+40}{8} = 5\text{ m}$
The centre of mass is located at $x = 5$ m — closer to the heavier mass (5 kg), as expected.
At 90°: $\tau_1 = rF\sin90° = (0.3)(20)(1) = 6\text{ N·m}$
At 30°: $\tau_2 = rF\sin30° = (0.3)(20)(0.5) = 3\text{ N·m}$
Torque is maximum (6 N·m) when the force is applied perpendicular to the wrench, and reduces to 3 N·m at 30° — illustrating why applying force perpendicular to a lever arm is most effective.
Weight $mg$ acts at the centre of mass, at distance $L/2$ from the pivot. Torque due to gravity: $\tau = mg\left(\dfrac{L}{2}\right) = (4)(10)(1) = 40\text{ N·m}$
Moment of inertia about the pivot: $I = \dfrac{1}{3}ML^2 = \dfrac{1}{3}(4)(2)^2 = \dfrac{1}{3}(4)(4) = \dfrac{16}{3}\text{ kg·m}^2$
Angular acceleration: $\alpha = \dfrac{\tau}{I} = \dfrac{40}{16/3} = 40\times\dfrac{3}{16} = 7.5\text{ rad/s}^2$
The angular acceleration just after release is 7.5 rad/s².
Acceleration while rolling: $a = \dfrac{g\sin\theta}{1 + I/MR^2}$
Solid cylinder: $I/MR^2 = 1/2$, so $a_{solid} = \dfrac{g\sin\theta}{1+0.5} = \dfrac{g\sin\theta}{1.5} = \dfrac{2g\sin\theta}{3}$
Hollow cylinder: $I/MR^2 = 1$, so $a_{hollow} = \dfrac{g\sin\theta}{1+1} = \dfrac{g\sin\theta}{2}$
Ratio: $\dfrac{a_{solid}}{a_{hollow}} = \dfrac{2/3}{1/2} = \dfrac{2}{3}\times2 = \mathbf{\dfrac{4}{3}}$
The solid cylinder accelerates faster (ratio 4:3), confirming it reaches the bottom first.
For rotational equilibrium (balance), the torques on both sides about the pivot must be equal:
$m_1g \times d_1 = m_2g \times d_2$ (the $g$ cancels)
$25 \times d_1 = 35 \times 1.5$
$25d_1 = 52.5 \Rightarrow d_1 = \dfrac{52.5}{25} = 2.1\text{ m}$
The 25 kg child must sit 2.1 m from the pivot for the see-saw to balance.
Rotational kinetic energy: $KE_{rot} = \dfrac{1}{2}I\omega^2$
Initial: $KE_1 = \frac{1}{2}(10)(10)^2 = \frac{1}{2}(10)(100) = 500\text{ J}$
Final: $KE_2 = \frac{1}{2}(10)(20)^2 = \frac{1}{2}(10)(400) = 2000\text{ J}$
Work required = $\Delta KE = 2000 - 500 = \mathbf{1500\text{ J}}$