Question

A solid cylinder attached with a horizontal massless spring (see figure) can roll without slipping along a horizontal surface. If the system is released from rest at a position in which spring is stretched by $0.3m.$
If $k=20N/m$, Rotational K.E. of cylinder as it passes through mean position is :

• A. $0.3J$
• B. $0.45J$
• C. $0.6J$
• D. $0.9J$

Answer 1

Answer: (A)

$0.3J$

Law of Conservation of Energy,

Translation energy = Rotational energy

$\frac{1}{2}k{x}^2=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$

$\frac{1}{2}\times 20\times {0.3}^2=\frac{1}{2}\times \frac{m{R}^2}{2}\times \frac{v^2}{R^2}+\frac{1}{2}m{v}^2$

$m{v}^2=1.2J$

kinetic energy of cylinder,when passes through the mean position

$\frac{m{v}^2}{4}=0.3J$