question_answer

A disc shaped body has two tight windings of light threads, one on the inner rim of radius \[R=1\text{ }m\]and the other on outer rim of radius 2R (see figure). It is kept on a horizontal surface and the ends of the two threads are pulled horizontally in opposite directions with force of equal magnitude\[F=20N\]. Mass of the body and its moment of inertia about an axis through centre 0 and perpendicular to the plane of the figure are \[M=4\text{ }kg\]and \[I=8\,kg{{m}^{2}}\] respectively. Find the kinetic energy of the body 2 seconds after the forces begin to act, if the surface is rough enough to ensure rolling without sliding.

   

A) \[\frac{200}{3}J\]         

B) \[\frac{100}{3}J\]

C) \[\frac{125}{3}J\]         

D) \[\frac{250}{9}J\]

Correct Answer: B

Solution :

[b] Let the friction be f. \[f=Ma\Rightarrow f=4a\]            ? (i) And \[I\alpha =F.2R-FR-f.2R\] \[\Rightarrow \,\,\,8.\alpha =20\times 1-2f\Rightarrow 4\alpha =10-f\] But   \[2R.\alpha =a\Rightarrow 2\alpha =a\] \[\therefore \,\,\,\,10-f=2a\]                ... (ii) Solving (i) and (ii),  \[a=\frac{10}{6}m/{{s}^{2}}=\frac{5}{3}m/{{s}^{2}}\] \[\alpha =\frac{a}{2R}=\frac{5}{6}rad/s\] Hence at \[t=2s;\] at \[V=at=\frac{10}{3}m/s\]and \[\omega =\alpha t=\frac{5}{3}rad/s\] \[\therefore \,\,\,\,K=\frac{1}{2}M{{V}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\times 4\times {{\left( \frac{10}{3} \right)}^{2}}+\frac{1}{2}\times 8\times {{\left( \frac{5}{3} \right)}^{2}}=\frac{100}{3}J\]