A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an

Question

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ______ and ______ respectively. [2026]

Smart Achievers · Accepted Answer

(4)

$ω_{i} = 1200 rpm = 1200 \times \frac{2 π}{60} = 40 π rad / \sec$

$ω_{f} = ω_{i} + α t \Rightarrow 40 π + α (10)$

$\Rightarrow α = - 4 π rad / \sec^{2}$

$I = \frac{2}{5} M R^{2} = \frac{2}{5} (5) {(0.04)}^{2} = 0.0032 kg - m^{2}$

$τ = I α = 0.0032 \times 4 π = 0.0128 π N - m$

$\Rightarrow τ = 0.0128 π N - m$

$ω_{f}^{2} = ω_{i}^{2} + 2 α θ$

$0 = {(40 π)}^{2} + 2 (4 π) θ \Rightarrow θ = 200 π rad$

$No. of revolution = \frac{θ}{2 π} = \frac{200 π}{2 π}$

$No. of revolution = 100$

Solution

1 0.128 $π$ Nm, 100

2 0.0128 $π$ Nm, 50

3 0.128 $π$ Nm, 50

4 0.0128 $π$ Nm, 100

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Solution

1 0.128π Nm, 100

2 0.0128π Nm, 50

3 0.128π Nm, 50

4 0.0128π Nm, 100

Ans. (4) ωi=1200 rpm=1200×2π60=40π rad/sec ωf=ωi+αt⇒40π+α(10) ⇒α=-4π rad/sec2 I=25MR2=25(5)(0.04)2=0.0032 kg-m2 τ=Iα=0.0032×4π=0.0128π N-m ⇒τ=0.0128π N-m ωf2=ωi2+2αθ 0=(40π)2+2(4π)θ⇒θ=200π rad No. of revolution=θ2π=200π2π No. of revolution=100

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1 0.128 $π$ Nm, 100

2 0.0128 $π$ Nm, 50

3 0.128 $π$ Nm, 50

4 0.0128 $π$ Nm, 100