Rotational Motion Question 181
Question: A wheel having angular momentum $2\pi kg-m^{2}/s$ about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in 30 sec would be
Options:
A) $\frac{\pi }{18}Nm$
B) $\frac{2\pi }{15}Nm$
C) $\frac{\pi }{12}Nm$
D) $\frac{\pi }{15}Nm$
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Answer:
Correct Answer: D
Solution:
[d]
$\tau \times \Delta t=L _{0}$
${\because \sin ceL _{f}=0}$
or $\tau \times 30=2\pi $
$\tau =\frac{\pi }{15}N-m$
BETA
