First slide

Rotational motion

Question

A horizontal 90 kg merry-go-round is a solid disk of radius 1.50 m and is started from rest by a constant horizontal force of 50.0 N applied tangentially to the edge of the disk. The kinetic energy of the disk after 3.00 is

Moderate

A

125 J
B

500 J
C

250 J
D

150 J

Solution

From the rigid object under a net torque model,
$\sum τ = I α \to α = \frac{\sum τ}{l} = \frac{F R}{\frac{1}{2} M R^{2}} = \frac{2 F}{M R}$
From the definition of rotational kinetic energy and the rigid object under constant angular acceleration model,
$\begin{array}{l} K = \frac{1}{2} I ω^{2} = \frac{1}{2} I {(ω_{i} + α t)}^{2} \\ = \frac{1}{2} I α^{2} t^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) {(\frac{2 F}{M R})}^{2} t^{2} = \frac{F^{2} t^{2}}{M} \\ Substituting, K = \frac{(50.0 N)^{2} (3.00 s)^{2}}{(90.0 kg)} = 250 J \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App