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QJEEM21Phab41: JEE Main 2021 Physics - Class 11 JEEM21Phab41 Solution: Consider a 20 kg uniform circular disk of radius 0.2 m. It is…

Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.

JEE Main 2021 (Online) 16th March Morning Shift Physics - Rotational Motion Question 111 English
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s −- 1.

The value of n, to the nearest integer, is __________.

[Given : In one complete revolution, the disk rotates by 6.28 rad]

Step-by-Step Verified Solution

α=τI=F.R.mR2/2=2FmR\alpha = {\tau \over I} = {{F.R.} \over {m{R^2}/2}} = {{2F} \over {mR}}

α=2×20020×(0.2)=10\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10 rad/s2

ω2=ω02+2αΔθ{\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta

(50)2=02+2(10)Δθ{(50)^2} = {0^2} + 2(10)\Delta \theta

⇒Δθ=250020\Rightarrow \Delta \theta = {{2500} \over {20}}

Δθ=125\Delta \theta = 125 rad

No. of revolution =1252π≈20= {{125} \over {2\pi }} \approx 20 revolution

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