Question 553: Centre of Mass and Rotational Motion

Question

A solid cylinder of mass $$3 \mathrm{~kg}$$ is rolling on a horizontal surface with velocity $$4 \mathrm{~m} / \mathrm{s}$$. It collides with a horizontal spring whose one end is fixed to rigid support. The force constant of material of spring is $$200 \mathrm{~N} / \mathrm{m}$$. The maximum compression produced in the spring will be (assume collision between cylinder & spring be elastic)

Accepted Answer

At maximum compression, the solid cylinder will stop.
So loss in K.E. of cylinder = Gain in P.E. of spring
$$\begin{array}{ll}
	herefore \quad & \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \mathrm{kx}^2 \
	herefore & \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \frac{\mathrm{mR}^2}{2}\left(\frac{\mathrm{v}}{\mathrm{R}}ight)^2=\frac{1}{2} \mathrm{kx}^2 \
	herefore \quad & \frac{3}{4} \mathrm{mv}^2=\frac{1}{2} \mathrm{kx}^2 \
	herefore \quad & \frac{3}{4} 	imes 3 	imes(4)^2=\frac{1}{2} 	imes 200 	imes \mathrm{x}^2 \
	herefore \quad & \frac{36}{100}=\mathrm{x}^2 \Rightarrow \mathrm{x}=0.6 \mathrm{~m}
\end{array}$$

Answer

0.7 m

Answer

0.2 m

Answer

0.5 m

Answer

0.6 m

Centre of Mass and Rotational Motion

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