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Q.

The wheel on a grinder is a uniform 2 kg disc of 25 cm radius. It comes uniformly to rest from 1440 rpm in a time of 31.4 s. How large a friction torque slows its motion?

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a

0.1 Nm

b

0.3 Nm

c

0.2 Nm

d

0.4 Nm

answer is B.

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Detailed Solution

ω=1440×2π60=48π rad/s2α=ωf−ωit=−48π31.4=−4.8 rad/s2I=12mr2=1220.252=62.5×10−3 kg.m2τ=Iα=62.5×10−3−4.8=0.3Nm
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