Rotational Motion Question 208
Question: A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions ${{s}^{-2}}$ is
Options:
A) 25 N
B) 50 N
C) 78.5 N
D) 157 N
Show Answer
Answer:
Correct Answer: D
Solution:
[d]
Here $\alpha =2revolutions/s^{2}=4\pi rad/s^{2}$
$ (given) $
$I _{cylinder}=\frac{1}{2}MR^{2}=\frac{1}{2}(50){{(0.5)}^{2}}$$ =$ $\frac{25}{4}Kg-m^{2}$
$ As $$\tau =I\alpha soTR=I\alpha $
$\Rightarrow T=\frac{I\alpha }{R}=\frac{\left( \frac{25}{4} \right)(4\pi )}{(0.5)}N=50\pi N=157N$
