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Question

Physics Question on Uniform Circular Motion

A solid body rotates an angle θ\theta about a stationary axis according to the law θ=6t2t3\theta = 6t - 2t^3 .What is the mean value of angular velocity over the time interval between t=0t = 0 and the time when the body comes to rest ?

A

1 rad/s

B

2 rad/s

C

3 rad/s

D

4 rad/s

Answer

4 rad/s

Explanation

Solution

Given, θ=6t2t3\theta=6t-2t^{3}...(i) dθdt=ω=66t2\therefore \frac{d\theta}{dt}=\omega=6-6t^{2} As, ω=0 \omega=0 (given) 66t2=06 - 6t^{2} = 0 t2=±1t^{2}=\pm1 As, t=1st =-1\, s is not possible, hence t=1st = 1 \,s. Substituting t=1st = 1\, s in E (i), we get θ1=62=4\theta_{1}=6-2=4 rad and θ0=6×02×02(t=0)=0\theta_{0}=6\times0-2\times0^{2}\left(t=0\right)=0 \therefore Mean value of ω=θ1+θ0t1+t0=4+01+0=4rads1\omega=\frac{\theta_{1}+\theta_{0}}{t_{1}+t_{0}}=\frac{4+0}{1+0}=4 rads^{-1}