Rotational Motion Question 80
Question: A solid cyclinder 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 $ 2rev/s^{2} $ is
Options:
A) 25 N
B) 50 N
C) 78.5 N
D) 157 N
Show Answer
Answer:
Correct Answer: D
Solution:
-
$ \Rightarrow $ $ m=50kg\Rightarrow r=0.5\Rightarrow \alpha =2rev/s^{2} $
$ \Rightarrow $ Torque produce by the tension in the string
$ =T\times R=T\times 0.5=\frac{T}{2}Nm $ …(i)
We know $ \tau =/\alpha $ ..(ii)
From Eqs.(i) and (ii),
$ =( \frac{MR^{2}}{2} )\times (2\times 2\pi )\frac{rad}{s^{2}} $
$ \therefore $ $ {l _{\text{solid cylinder}}}=\frac{MR^{2}}{2} $
$ \frac{T}{2}=\frac{50\times {{(0.5)}^{2}}}{2}\times 4\pi $
$ T=50\times \frac{1}{4}\times 4\pi $
$ =50\pi =157N $
