First slide

Dynamics of rotational motion about fixed axis

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 s is

Moderate

A

125 J
B

500 J
C

250 J
D

150 J

Solution

From the rigid object under a net torque model,
$\sum_{} τ = lα \overset{}{\to α = \frac{\sum_{} τ}{l} = \frac{FR}{\frac{1}{2} {MR}^{2}} = \frac{2 F}{MR}}$
From the definition of rotational kinetic energy and the rigid object under constant angular acceleration model.
$K = \frac{1}{2} {Iω}^{2} = \frac{1}{2} I {(ω_{1} + αt)}^{2}$
= $\frac{1}{2} {Iα}^{2} t^{2} = (\frac{1}{2} {MR}^{2}) {(\frac{2 F}{MR})}^{2} t^{2} = \frac{F^{2} t^{2}}{M}$
Substituting, K = $\frac{{(50.0 N)}^{2} {(3.00 s)}^{2}}{(90.0 kg)} = 250 J$

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