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Question: A grindstone has a moment of inertia of\(1.6\times {{10}^{-3}}kg{{m}^{2}}\). When a constant torque ...

A grindstone has a moment of inertia of1.6×10−3kgm21.6\times {{10}^{-3}}kg{{m}^{2}}. When a constant torque is applied, the flywheel reaches an angular velocity of 1200rev/min in 15s. Assuming it started from rest, Find:
A. the angular acceleration
B. The unbalanced torque applied
C. The angle turned through in the 15s
D. the work W done on the flywheel by the torque

Explanation

Solution

We are given certain quantities and we could first correctly note them down. Then we could recall the expressions for each of the physical quantities related to the flywheel to be found. Then, we could simply substitute the known values into those expressions and hence find the answer.
Formula used:
Angular acceleration,
α=ωt\alpha =\dfrac{\omega }{t}
Torque,
τ=Iα\tau =I\alpha
Angle,
θ=ω22α\theta =\dfrac{{{\omega }^{2}}}{2\alpha }
Work done,
W=Iω22W=\dfrac{I{{\omega }^{2}}}{2}

Complete step by step answer:
In the question, we are given a grindstone of moment of inertia1.6×10−3kgm21.6\times {{10}^{-3}}kg{{m}^{2}}. We are also said that the flywheel reaches an angular velocity of 1200rev/min in 15s on applying a constant torque. We are also asked to assume it to start from rest and hence find certain quantities.
A. Angular acceleration,
We know that angular acceleration can be given by,
α=ωt\alpha =\dfrac{\omega }{t}
Where, ω\omega is the angular velocity and t is the time taken.
ω=1200×2πrad/min⁡=40πrad/sec⁡\omega =1200\times 2\pi rad/\min =40\pi rad/\sec
So,
α=40π15rad/sec⁡2=8π3rad/sec⁡2\alpha =\dfrac{40\pi }{15}rad/{{\sec }^{2}}=\dfrac{8\pi }{3}rad/{{\sec }^{2}}
So we found the angular acceleration to be 8π3rad/sec⁡2\dfrac{8\pi }{3}rad/{{\sec }^{2}}
B. Torque,
τ=Iα\tau =I\alpha
Substituting values,
τ=1.6×10−3×8π3=12.8π3×10−3Nm\tau =1.6\times {{10}^{-3}}\times \dfrac{8\pi }{3}=\dfrac{12.8\pi }{3}\times {{10}^{-3}}Nm
So we found the torque to be 12.8π3×10−3Nm\dfrac{12.8\pi }{3}\times {{10}^{-3}}Nm
C. The angle turned through in the 15s,
θ=ω22α=(40π)2×32×12.8×π=4.68πrad\theta =\dfrac{{{\omega }^{2}}}{2\alpha }=\dfrac{{{\left( 40\pi \right)}^{2}}\times 3}{2\times 12.8\times \pi }=4.68\pi rad
D. the work W done on the flywheel by the torque,
W=Iω22=1.6×10−3×(40π)22=0.64π2JW=\dfrac{I{{\omega }^{2}}}{2}=\dfrac{1.6\times {{10}^{-3}}\times {{\left( 40\pi \right)}^{2}}}{2}=0.64{{\pi }^{2}}J

So, we found the work done on the flywheel to be0.64π2J0.64{{\pi }^{2}}J.

Note: In the question we are given the angular velocity in units of revolutions per minute which is not the SI units. Since all the other quantities are given in SI units we will convert this into radians per second which is the SI unit for angular velocity. Doing so will get you all the final answers in the SI unit.