Rotational Motion

JEE Main level, numerical answer, medium

166 of 280

Question 166 of 280

Consider a 20kg20kg uniform circular disk of radius 0.2m0.2m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F=20NF=20N through a massless string wrapped around its periphery as shown in the figure.

Suppose the disk makes nn number of revolutions to attain an angular speed of 50rads−150rad{s}^{-1}. The value of nn, to the nearest integer, is _______ .
[Given : In one complete revolution, the disk rotates by 6.28rad6.28rad]

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Answer: 20

α=τI=F.R.mR2/2=2FmR\alpha =\dfrac{\tau }{I}=\dfrac{F.R.}{m{R}^{2}/2}=\dfrac{2F}{mR}
α=2×20020×(0.2)=10rad/s2\alpha =\dfrac{2\times 200}{20\times \left(0.2\right)}=10rad/{s}^{2}
ω2=ω02+2αΔθ{\omega }^{2}={\omega }_{0}^{2}+2\alpha \Delta \theta
(50)2=02+2(10)Δθ⇒Δθ=250020{\left(50\right)}^{2}={0}^{2}+2\left(10\right)\Delta \theta \Rightarrow \Delta \theta =\dfrac{2500}{20}
Δθ=125rad\Delta \theta =125rad
No. of =1252π≈20=\dfrac{125}{2\pi }\approx 20 revolution

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