Centripetal Acceleration — Washing Machine Spin (Worked Example)

Hard

Problem

A washing machine drum of radius 0.25 m spins at 1200 rpm during the spin cycle. What is the centripetal acceleration of clothes stuck to the drum wall?

Given

r = 0.25 m
1200 rev/min

Find

Centripetal acceleration a.

Principle

Convert rpm to tip speed first (v = ωr), then a = v²/r. The wall pushes the clothes inward — that push is what squeezes water out.

Formula

a = \frac{v^2}{r},\quad v = \omega r

a = v²/r with v = ωr

Solution

Convert: $\omega = 1200 \times 2\pi/60 = 125.7\ \mathrm{rad/s}$
Tip speed: $v = \omega r = 125.7 \times 0.25 = 31.4\ \mathrm{m/s}$
Accelerate: $a = v^2/r = 31.4^2/0.25$
Calculate: $a = 986/0.25 = 3944\ \mathrm{m/s^2}$

3.94 × 10³ m/s² (≈ 402g) toward the drum centre

Unit check

(m/s)² ÷ m = m/s² — correct.

What this means physically

About 400 times Earth's gravity — water is flung through the holes while fabric, pressed by the wall, stays. Spin cycles are applied physics: extreme centripetal acceleration as a drying machine.

Common mistake

Using the diameter as r, or skipping the rpm→rad/s conversion. Both errors change the answer by factors of 2–60 — convert carefully.