Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A uniform solid cylinder of mass M and radius R rotates about a frictionless axle. Two masses are suspended from a rope wrapped around the cylinder. If the system is released from rest, the tension in each rope is

\[\begin {array} {1 1} (a)\;\frac{Mmg}{M+m} & \quad (b)\;\frac{Mmg}{M+2m} \\ (c)\;\frac{Mmg}{M+3m} & \quad  (d)\;\frac{Mmg}{M+4m} \end {array}\]

Can you answer this question?

1 Answer

0 votes
$mg-T=ma$. for each man.
$2TR= I \alpha$
$\qquad= \large\frac{MR^2}{2}. \alpha$
$T= \large\frac{MR \alpha}{4}$
Since string does not stop,
$a= R \alpha$
$\therefore mg= ma +\large\frac{Ma}{4}$
$\therefore a= \large\frac{4\;mg}{4m+M}$
$T= \large\frac{M}{4} \frac{4mg}{4m+M}$
$T= \large\frac{Mmg}{(M+4m)}$
answered Nov 27, 2013 by meena.p
edited Jun 19, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App