A solid cylinder of mass 3, kg is rolling on | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,At\,{m{maximum}}\,compression\,the\
solid\,cylinder\,will\,stop.\
According\,to\,law\,of\,conservation\

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,At\,{m{maximum}}\,compression\,the\
solid\,cylinder\,will\,stop.\
According\,to\,law\,of\,conservation\,of\,\
mechanical\,energy\
Loss\,in\,kinetic\,energy\, = Gain\,in\,potential\
energy\,of\,cylinder\,of\,spring\
\frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = \frac{1}{2}k{x^2}\
\frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{{m{R^2}}}{2}} ight){\left( {\frac{v}{R}} ight)^2} = \frac{1}{2}k{x^2}
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,(v = R\omega \,and\,for\,solid\,cylinder,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = \frac{1}{2}m{R^2})\
\frac{1}{2}m{v^2} = \frac{1}{4}m{v^2} = \frac{1}{2}k{x^2}\
\frac{3}{4}m{v^2} = \frac{1}{2}k{x^2}\,or\,{x^2} = \frac{3}{2}\,\frac{{m{v^2}}}{k}\
Here,\,m = 3\,kg,\,v = 4\,m\,{s^{ - 1}},\,k = 200\,N\,{m^{ - 1}}\
Subsitituting\,the\,given\,values,\,we\,get\
{x^2} = \frac{{3 	imes 3 	imes 4 	imes 4}}{{2 	imes 200}} \Rightarrow {x^2} = \frac{{36}}{{100}}\,or\,x = 0.6\,m
\end{array}$

A solid cylinder of mass $3\, kg$ is rolling on a horizontal surface with velocity $4\, m s^{- 1}$. It collides with a horizontal spring of force constant $200 \,N m^{-1}$. The maximum compression produced in the spring will be ............... $\mathrm{m}$

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