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A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 $ms^{-1}$. It collides with a horizontal spring of force constant 200 $Nm^{-1}$. The maximum compression produced in the spring will be
A
0.2 m
B
0.5 m
C
0.6 m
D
0.7 m
Detailed Solution
At maximum compression all the kinetic energy (translational + rotational) is stored in the spring:
$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}kx^2$, with $I = \frac{mR^2}{2}$ and $\omega = \frac{v}{R}$
$\frac{1}{2}\times3\times(4)^2 + \frac{1}{2}\times\frac{3R^2}{2}\times\left(\frac{4}{R}\right)^2 = \frac{1}{2}\times200\times x^2$
$24 + 12 = 100x^2 \Rightarrow x^2 = 0.36$
$x = 0.6$ m
$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}kx^2$, with $I = \frac{mR^2}{2}$ and $\omega = \frac{v}{R}$
$\frac{1}{2}\times3\times(4)^2 + \frac{1}{2}\times\frac{3R^2}{2}\times\left(\frac{4}{R}\right)^2 = \frac{1}{2}\times200\times x^2$
$24 + 12 = 100x^2 \Rightarrow x^2 = 0.36$
$x = 0.6$ m