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  • A ring of radius '  ' rotating we th an angular velocity of ' <img alt="2 \

A ring of radius ' 2 \mathrm{~m} ' rotating we th an angular velocity of ' 2 \mathrm{rad} / \mathrm{s} ' is placed on a rough horizontal surface. A translational velocity of ' 2 \mathrm{~m} / \mathrm{s} ' l's given to the ring. What will be the final translatory velocity of the ring after it starts pure rolling?

Option: 1

1 \mathrm{~m} / \mathrm{s}


Option: 2

2 \mathrm{~m} / \mathrm{s}


Option: 3

0.5 \mathrm{~m} / \mathrm{s}


Option: 4

1.5 \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

\begin{aligned} & v=u-\left(\frac{f}{m}\right)t \\ & \omega=\omega_0-\left(\frac{f t}{m r}\right) \end{aligned}

using the 2nd law of newton:

m(v-u)=ft

using the work energy theorem
\begin{aligned} & \frac{1}{2} m u^2+\frac{1}{2} m \omega_0^2 r^2-\frac{1}{2} m v^2-\frac{1}{2} m \omega^2 r^2= \\ & f\left[\frac{\left(u+\omega_0 r\right)+(v+\omega r)}{2}\right] t \end{aligned}

As, u<\omega_0 \gamma

In case of pure rolling,

\begin{aligned} & 2 v=\omega_0 r-u \\ & v=\frac{\omega_0 r-u}{2} \\ & v=\frac{2 \times 2-2}{2}=1 \mathrm{~m} / \mathrm{s} \end{aligned}

 

Posted by

vinayak

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