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Question

A wheel having a moment of inertia 2kg-m2 about its vertical axis rotates at the rate of 60rpm about this axis. The torque which can stop the wheel's rotation in one minute would be


A

2π13N-m

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B

π14N-m

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C

π15N-m

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D

π20N-m

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Solution

The correct option is C

π15N-m


Step 1: Given parameters

Moment of inertia, I = 2kg-m2.

Rotating rate, ω0 = 60rpm

ω0=6060×2πrads

Time, t=60s

Step 2: Formula used

The torque required to stop the wheel's rotation is given as follows:

τ==lω0-ωtα=ω0-ωt

Where, I is the moment of inertia,τ is the torque, t is the time, ω0 is the initial angular velocity, ω is the final angular velocity, α is the retardation

As the final angular velocity is zero,

α=ω0-0tα=ω0t

Therefore, the above formula becomes,

τ=I×ατ=I×ω0t...1

Step 3: Calculating the value of torque

Substitute the given values in equation (1).

τ=I×ω0tτ=2×2π×6060×60

τ=π15N-m

Hence, the value of torque is π15N-m.

Hence, option (C) is the correct answer.


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