Question

A semicircular track of radius R=62.5 cm is cut in a block .Mass of block ,having track,is M=1 kg and rests over a smooth horizontal floor.A cylinder of radius r=10cm and mass m=0.5 kg is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure.When the thread is burnt ,cylinder starts to move down the track.Sufficient friction exists between surface of cylinder and track , so that cylinder does not slip. Calculate velocity of axis of cylinder and velocity of the block when it reaches bottom of the track.Also find force applied by block on the floor at that moment.(g=10 $m/s^2$)

Answer 1

By momentum conservation $mu+mv=0$

$u=\frac{m}{m}v=\frac{v}{2}$

Change in potential energy $=$ total kinetic energy

$mg(R-r)=\frac{1}{2}m{\omega }^2+\frac{1}{2}M{\left(\frac{v}{2}\right)}^2$

$mg(R-r)=\frac{1}{2}m{\omega }^2+\frac{1}{2}M{\left(\frac{v}{2}\right)}^2\left(1\right)0.5\times 10(62.5\times 10)\times {10}^2=\frac{1}{2}\frac{m{v}^2{\omega }^2}{2}+\frac{1}{2}\frac{m{v}^2}{4}0.5\left(52.5\right){10}^{-1}=\frac{1}{4}(\frac{1}{2}{v}^2+\frac{1}{2}{v}^2)\frac{1}{2}\left(5.25\right)=\frac{1}{4}{v}^2{v}^2=10.5v=\sqrt{10.5}$

(2) speed of block $=-\frac{v}{2}=\frac{\sqrt{10.5}}{2}=\sqrt{2.625}m$

(3) Force applied by cylinder

$=mg+\frac{m{v}^2}{R}=0.5\times 10+\frac{0.5\times 10.5}{62.5\times {10}^{-2}}=5+8.4=13.5Ndownwards$