System of Particles and Rotational Motion • Topic 3 of 3

Rotational Dynamics & Rolling

Just as $F=ma$ governs straight-line motion, an exactly parallel law governs rotation. A net torque produces angular acceleration in direct proportion to the torque and inversely to the moment of inertia. This makes rotational dynamics a near-perfect mirror of translational dynamics — every linear quantity has a rotational twin.

The rotational equation of motion is:

  • $\tau=I\alpha$, where $\alpha$ is the angular acceleration ($\text{rad/s}^2$) — the rotational analogue of $F=ma$.
  • The work done by a torque through an angle $\theta$ is $W=\tau\theta$, and the power delivered is $P=\tau\omega$.
  • Rotational kinetic energy: a body rotating with angular speed $\omega$ stores energy $KE_{rot}=\frac{1}{2}I\omega^2$, the analogue of $\frac{1}{2}mv^2$.
  • Work–energy theorem for rotation: $W_{net}=\Delta KE_{rot}=\frac{1}{2}I\omega_2^2-\frac{1}{2}I\omega_1^2$.

Rolling without slipping combines translation of the centre of mass with rotation about it. The key condition is that the contact point is instantaneously at rest, which links the two motions:

  • $v=\omega R$ (velocity of the centre) and $a=\alpha R$ (acceleration), where $R$ is the radius.
  • The total kinetic energy is the sum of translational and rotational parts: $KE=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$.
  • Using $I=mk^2$ and $v=\omega R$, this becomes $KE=\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)$.

Rolling down an incline. When a body rolls (without slipping) down a slope of height $h$, the lost potential energy $mgh$ becomes shared between translational and rotational KE. Solving energy conservation gives the speed at the bottom and the acceleration along the incline:

  • $v=\sqrt{\dfrac{2gh}{1+\frac{k^2}{R^2}}}$ and $a=\dfrac{g\sin\theta}{1+\frac{k^2}{R^2}}$.
  • Because $\frac{k^2}{R^2}$ is smallest for a sphere ($\tfrac{2}{5}$), larger for a disc ($\tfrac{1}{2}$), and largest for a ring ($1$), a solid sphere reaches the bottom first, then the disc, then the ring — regardless of mass or radius.
A disc rolling without slipping down an inclinev = ωRωθhRolling: contact point at rest, v = ωR, KE = ½mv² + ½Iω²
Solved Examples
1
Worked Example
A torque of $6\ \text{N m}$ acts on a flywheel of moment of inertia $3\ \text{kg m}^2$. Find the angular acceleration.
Solution
  1. Step 1: Use $\tau=I\alpha$, so $\alpha=\frac{\tau}{I}$.
  2. Step 2: Substitute: $\alpha=\frac{6}{3}$.
  3. Step 3: Compute: $\alpha=2\ \text{rad/s}^2$.

Answer: $\alpha=2\ \text{rad/s}^2$.

2
Worked Example
A disc of moment of inertia $0.5\ \text{kg m}^2$ rotates at $4\ \text{rad/s}$. Find its rotational kinetic energy.
Solution
  1. Step 1: Use $KE_{rot}=\frac{1}{2}I\omega^2$.
  2. Step 2: Substitute: $KE=\frac{1}{2}(0.5)(4)^2=\frac{1}{2}(0.5)(16)$.
  3. Step 3: Compute: $KE=4\ \text{J}$.

Answer: $KE_{rot}=4\ \text{J}$.

3
Worked Example
A wheel of radius 0.4 m rolls without slipping with a centre speed of $6\ \text{m/s}$. Find its angular speed.
Solution
  1. Step 1: For rolling without slipping $v=\omega R$, so $\omega=\frac{v}{R}$.
  2. Step 2: Substitute: $\omega=\frac{6}{0.4}$.
  3. Step 3: Compute: $\omega=15\ \text{rad/s}$.

Answer: $\omega=15\ \text{rad/s}$.

4
Worked Example
A solid sphere of mass 2 kg rolls without slipping at $3\ \text{m/s}$. Find its total kinetic energy. (For a solid sphere $\frac{k^2}{R^2}=\frac{2}{5}$.)
Solution
  1. Step 1: Use $KE=\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)$.
  2. Step 2: Substitute: $KE=\frac{1}{2}(2)(3)^2\left(1+\frac{2}{5}\right)=9\times\frac{7}{5}$.
  3. Step 3: Compute: $KE=12.6\ \text{J}$.

Answer: $KE=12.6\ \text{J}$.

5
Worked Example
A disc rolls without slipping from rest down an incline of height 2.1 m. Find the speed at the bottom. (For a disc $\frac{k^2}{R^2}=\frac{1}{2}$, $g=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Use $v=\sqrt{\dfrac{2gh}{1+\frac{k^2}{R^2}}}$.
  2. Step 2: Substitute: $v=\sqrt{\dfrac{2(9.8)(2.1)}{1+\frac{1}{2}}}=\sqrt{\dfrac{41.16}{1.5}}=\sqrt{27.44}$.
  3. Step 3: Compute: $v\approx5.24\ \text{m/s}$.

Answer: $v\approx5.24\ \text{m/s}$.

6
Worked Example
A ring and a solid sphere of the same mass and radius roll down the same incline from rest. Which reaches the bottom first, and why?
Solution
  1. Step 1: Acceleration is $a=\dfrac{g\sin\theta}{1+\frac{k^2}{R^2}}$; a larger $\frac{k^2}{R^2}$ gives a smaller acceleration.
  2. Step 2: For the ring $\frac{k^2}{R^2}=1$; for the solid sphere $\frac{k^2}{R^2}=\frac{2}{5}$.
  3. Step 3: The sphere has the smaller $\frac{k^2}{R^2}$, so the larger acceleration, independent of mass and radius.

Answer: The solid sphere reaches the bottom first.

Key Points

  • Rotational Newton's law: $\tau=I\alpha$; work $W=\tau\theta$ and power $P=\tau\omega$.
  • Rotational kinetic energy: $KE_{rot}=\frac{1}{2}I\omega^2$.
  • Rolling without slipping requires $v=\omega R$ and $a=\alpha R$.
  • Total KE while rolling: $KE=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)$.
  • Down an incline $a=\dfrac{g\sin\theta}{1+k^2/R^2}$; the solid sphere beats the disc, which beats the ring.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The rotational analogue of $F=ma$ is:
Explanation: Net torque equals moment of inertia times angular acceleration: $\tau=I\alpha$.
Q2.The rotational kinetic energy of a body is:
Explanation: Rotational KE is $\frac{1}{2}I\omega^2$, mirroring $\frac{1}{2}mv^2$.
Q3.For a wheel rolling without slipping, the velocity of the centre is:
Explanation: The rolling condition (contact point at rest) gives $v=\omega R$.
Q4.A solid sphere, a disc and a ring roll down the same incline from rest. The first to reach the bottom is the:
Explanation: The sphere has the smallest $k^2/R^2$ ($\tfrac{2}{5}$), so the largest acceleration, independent of mass and radius.
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