System of Particles and Rotational Motion • Topic 2 of 3

Torque, Moment of Inertia & Angular Momentum

Force makes things move; torque makes things turn. The turning effect of a force about an axis is called torque (or moment of force). It depends not just on how hard you push, but on where and in what direction you push. That is why a door opens easily at the handle but barely budges near the hinge.

Torque is the cross product of the position vector $\vec{r}$ (from the axis to the point of application) and the force $\vec{F}$:

  • $\vec{\tau}=\vec{r}\times\vec{F}$, with magnitude $\tau=rF\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.
  • $\tau=F\times(r\sin\theta)=F\times d$, where $d=r\sin\theta$ is the lever arm (perpendicular distance from the axis to the line of force).
  • The SI unit of torque is the newton metre ($\text{N m}$); its direction is along the axis of rotation (right-hand rule).

Moment of inertia is the rotational analogue of mass — it measures a body's resistance to angular acceleration. For a system of particles it is the mass-weighted sum of the squares of distances from the axis:

  • $I=\sum m_i r_i^2$ (continuous body: $I=\int r^2\,dm$). SI unit: $\text{kg m}^2$.
  • It depends on the total mass and how that mass is distributed about the axis — mass far from the axis contributes much more.
  • Standard results: thin ring/hoop about its axis $I=MR^2$; solid disc/cylinder about its axis $I=\tfrac{1}{2}MR^2$; solid sphere about a diameter $I=\tfrac{2}{5}MR^2$; thin rod about its centre $I=\tfrac{1}{12}ML^2$.

Radius of gyration $k$ is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia: $I=Mk^2$, so $k=\sqrt{I/M}$.

Angular momentum is the rotational analogue of linear momentum. For a particle $\vec{L}=\vec{r}\times\vec{p}$, and for a rigid body rotating about a fixed axis $L=I\omega$ (SI unit $\text{kg m}^2\text{/s}$). When the net external torque is zero, $\frac{d\vec{L}}{dt}=\vec{\tau}_{ext}=0$, so angular momentum is conserved: $I_1\omega_1=I_2\omega_2$. An ice-skater spins faster by pulling the arms in (reducing $I$, raising $\omega$) — a direct demonstration of this law.

Torque: position vector r, force F and lever arm daxis OrFd = r sin θθTorque τ = r F sinθ = F·d (d = perpendicular lever arm)
Solved Examples
1
Worked Example
A force of 20 N is applied at the end of a wrench 0.25 m long, perpendicular to it. Find the torque about the bolt.
Solution
  1. Step 1: Use $\tau=rF\sin\theta$ with $\theta=90^\circ$, so $\sin\theta=1$.
  2. Step 2: Substitute: $\tau=0.25\times20\times1$.
  3. Step 3: Compute: $\tau=5\ \text{N m}$.

Answer: $\tau=5\ \text{N m}$.

2
Worked Example
A force of 50 N acts at a point 0.4 m from the axis, making an angle of $30^\circ$ with the position vector. Find the torque.
Solution
  1. Step 1: $\tau=rF\sin\theta$.
  2. Step 2: Substitute: $\tau=0.4\times50\times\sin30^\circ=0.4\times50\times0.5$.
  3. Step 3: Compute: $\tau=10\ \text{N m}$.

Answer: $\tau=10\ \text{N m}$.

3
Worked Example
Two point masses of 2 kg each are fixed at the ends of a light rod of length 2 m. Find the moment of inertia about an axis through the centre, perpendicular to the rod.
Solution
  1. Step 1: Each mass is at $r=1$ m from the centre.
  2. Step 2: $I=\sum m_i r_i^2=2(1)^2+2(1)^2$.
  3. Step 3: Compute: $I=2+2=4\ \text{kg m}^2$.

Answer: $I=4\ \text{kg m}^2$.

4
Worked Example
A solid disc has mass 4 kg and radius 0.5 m. Find its moment of inertia about the central axis, and its radius of gyration.
Solution
  1. Step 1: For a disc $I=\tfrac{1}{2}MR^2=\tfrac{1}{2}(4)(0.5)^2$.
  2. Step 2: Compute: $I=\tfrac{1}{2}(4)(0.25)=0.5\ \text{kg m}^2$.
  3. Step 3: Radius of gyration $k=\sqrt{I/M}=\sqrt{0.5/4}=\sqrt{0.125}$.

Answer: $I=0.5\ \text{kg m}^2$ and $k\approx0.354\ \text{m}$.

5
Worked Example
A disc of moment of inertia $0.2\ \text{kg m}^2$ rotates at an angular speed of $10\ \text{rad/s}$. Find its angular momentum.
Solution
  1. Step 1: Use $L=I\omega$.
  2. Step 2: Substitute: $L=0.2\times10$.
  3. Step 3: Compute: $L=2\ \text{kg m}^2\text{/s}$.

Answer: $L=2\ \text{kg m}^2\text{/s}$.

6
Worked Example
A skater spinning at $2\ \text{rev/s}$ with moment of inertia $4\ \text{kg m}^2$ pulls in the arms, reducing the moment of inertia to $1\ \text{kg m}^2$. Find the new rate of spin.
Solution
  1. Step 1: No external torque, so angular momentum is conserved: $I_1\omega_1=I_2\omega_2$.
  2. Step 2: $\omega_2=\frac{I_1\omega_1}{I_2}=\frac{4\times2}{1}$.
  3. Step 3: Compute: $\omega_2=8\ \text{rev/s}$.

Answer: The skater spins at $8\ \text{rev/s}$ (four times faster).

Key Points

  • Torque is the turning effect of a force: $\vec{\tau}=\vec{r}\times\vec{F}$, magnitude $\tau=rF\sin\theta=Fd$.
  • Moment of inertia is rotational mass: $I=\sum m_i r_i^2$; it depends on mass distribution about the axis.
  • Standard results: ring $MR^2$, disc $\tfrac{1}{2}MR^2$, solid sphere $\tfrac{2}{5}MR^2$, rod (centre) $\tfrac{1}{12}ML^2$.
  • Radius of gyration: $I=Mk^2$, so $k=\sqrt{I/M}$.
  • Angular momentum $L=I\omega$; if net external torque is zero, $I_1\omega_1=I_2\omega_2$ (conservation).
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The torque produced by a force is maximum when the angle between $\vec{r}$ and $\vec{F}$ is:
Explanation: $\tau=rF\sin\theta$ is maximum when $\sin\theta=1$, i.e. $\theta=90^\circ$.
Q2.The moment of inertia of a thin ring of mass $M$ and radius $R$ about its central axis is:
Explanation: All the mass of a ring lies at distance $R$ from the axis, so $I=MR^2$.
Q3.The rotational analogue of linear momentum is:
Explanation: Angular momentum $L=I\omega$ corresponds to linear momentum $p=mv$.
Q4.An ice-skater pulls in the arms while spinning. The angular speed:
Explanation: With no external torque, $L=I\omega$ is constant; reducing $I$ increases $\omega$.
Practice more MCQs → 📝 Assignment · 30 marks