A wheel with moment of inertia 10 kg `m^(2)` is rotating at `(2//pi)` rps on its axis. The frictional torque acting on it, if it makes 20 rotations be
A wheel with moment of inertia 10 kg `m^(2)` is rotating at `(2//pi)` rps on its axis. The frictional torque acting on it, if it makes 20 rotations befor stopping is
A. `pi Nm`
B. `1//pi Nm`
C. `2 pi Nm`
D. `2//pi Nm`
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Correct Answer - D
`omega_(2)^(2)-omega_(1)^(2)=2 prop theta`
`therefore 4pi^(2)(0-(4)/(pi^(2)))=2xx alpha xx 2pi N`
`16= 4xx pi xx 20xx alpha`
`therefore alpha =(16)/(80xxpi)`
`therefore tau = I alpha =10xx(1)/(5pi)=(2)/(pi)Nm`.
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