Rotating disc: rim speed

Problem

A disc of radius $R = 0{,}40\;\mathrm{m}$ rotates with angular velocity $\omega = 50\;\mathrm{rad/s}$. Calculate the speed of a point on the rim and its centripetal acceleration.

Solution

The linear (tangential) speed of a point on the rim is related to the angular velocity by $v = \omega R$: $v = \omega R = 50 \cdot 0{,}40 = 20\;\mathrm{m/s}$ The centripetal acceleration of a point in uniform circular motion is $a_c = \omega^2 R$ (equivalently $a_c = v^2/R = \omega v$): $a_c = \omega^2 R = 50^2 \cdot 0{,}40 = 2500 \cdot 0{,}40 = 1000\;\mathrm{m/s^2}$ Comparing with $g = 9{,}81\;\mathrm{m/s^2}$: $\frac{a_c}{g} = \frac{1000}{9{,}81} \approx 102 \approx 100$ that is, a point on the rim experiences a centripetal acceleration of about 100 times that of gravity. $\ev{v = \omega R = 20\;\mathrm{m/s}, \qquad a_c = \omega^2 R = 1000\;\mathrm{m/s^2} \approx 100\,g}$

Linked items

Topics: Dinamica Concepts: Velocità angolare · Accelerazione centripeta · Moto circolare uniforme Objects: Disco rotante