---
title: "Man walking on a rotating platform"
subject: "Engineering Mechanics"
chapter: 8
key: "semester_3/engineering_mechanics:rotating-platform-man-walks"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rotating-platform-man-walks
site: BCT Engineering
---

# Man walking on a rotating platform

Engineering Mechanics · Chapter 8: Plane Motion of Rigid Bodies: Impulse and Momentum Method

- Asked in: year not given, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rotating-platform-man-walks
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-8

## Question

A horizontal circular platform, a uniform disc of mass 40 kg and radius 1.5 m, rotates freely about its vertical central axis at 20 rpm. A man of mass 60 kg standing at the centre walks slowly to the rim. Treating the man as a particle, find the new angular speed of the platform and the change in kinetic energy. Account for the change.

## Short answer

By angular momentum conservation, with initial moment of inertia 45 kg·m² and final 180 kg·m², the new speed is 5.0 rpm. Kinetic energy decreases by 74.0 J because the man does internal work while moving outward.

## Answer

**Given:** platform $M = 40$ kg, $R = 1.5$ m (disc), man $m = 60$ kg, initial speed $N_1 = 20$ rpm. No external torque acts about the vertical axis (the platform rotates freely), so **angular momentum is conserved**.

#### Moments of inertia
- Platform: $I_p = \tfrac12 MR^2 = \tfrac12(40)(1.5)^2 = 45.0$ kg·m²
- Man at the centre: contributes $\approx 0$. Initial $I_1 = 45.0$ kg·m².
- Man at the rim: $mR^2 = 60(1.5)^2 = 135$ kg·m². Final $I_2 = 45 + 135 = 180.0$ kg·m².

#### Final angular speed

$$
I_1\omega_1 = I_2\omega_2 \;\Rightarrow\; N_2 = N_1\frac{I_1}{I_2} = 20\times\frac{45}{180} = 5.00\ \text{rpm}
$$

In rad/s: $\omega_1 = 2.094$ and $\omega_2 = 0.524$ rad/s.

#### Change in kinetic energy

$$
T_1 = \tfrac12 I_1\omega_1^2 = \tfrac12(45)(2.094)^2 = 98.7\ \text{J}
$$

$$
T_2 = \tfrac12 I_2\omega_2^2 = \tfrac12(180)(0.524)^2 = 24.7\ \text{J}
$$

$$
\Delta T = T_2 - T_1 = -74.0\ \text{J}
$$

#### Reason
Angular momentum is conserved, but kinetic energy is not, because the man does internal work (muscular effort and friction between his feet and the platform) while he moves radially outward. The system's energy decreases by 74.0 J; that energy is dissipated, mainly as heat in the man's body and at the feet. Since $T = \tfrac12 I\omega^2 = H^2/2I$, an increase in $I$ at constant $H$ always reduces $T$.

**Answer: New speed = 5.0 rpm; kinetic energy decreases by 74.0 J.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is angular momentum conserved as the man walks outward?
   A. Kinetic energy is constant
   B. No external torque acts about the vertical axis
   C. The man's mass is small
   D. The platform is a uniform disc
   Answer: B. Platform rotates freely and the man's push is internal. (Key point: Conserved quantity)

2. What is I with the man at the rim?
   A. 45 kg·m^2
   B. 135 kg·m^2
   C. 180 kg·m^2
   D. 90 kg·m^2
   Answer: C. I = (1/2)(40)(1.5)^2 + 60(1.5)^2 = 45 + 135 = 180 kg·m^2. (Key point: Final moment of inertia)

3. What is the new speed?
   A. 5 rpm
   B. 10 rpm
   C. 20 rpm
   D. 80 rpm
   Answer: A. N2 = 20 × 45/180 = 5 rpm. (Key point: Final speed)

4. Why does kinetic energy fall by 74 J although angular momentum is conserved?
   A. Energy is destroyed
   B. Since T = H^2/2I, I increases at constant H so T decreases; the difference is internal work dissipated as heat
   C. Friction with the axle removes it
   D. The platform slows because the man adds mass
   Answer: B. At fixed H, larger I means smaller T, and the man's muscular effort and foot friction dissipate it. (Key point: Energy change)

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