Questions
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?
detailed solution
Correct option is B
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=0.3NmTalk to our academic expert!
Similar Questions
Forces are applied on a wheel of radius 20 cm as shown in the figure. The torque produced by the forces 8 N at A, 8 N at E, 6 N at C and 9 N at D at angles indicated is
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