Question 423: Centre of Mass and Rotational Motion

Question

A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $$3 \mathrm{~m} / \mathrm{s}$$ (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.

Accepted Answer

Total initial kinetic energy
$$
=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \mathrm{I} \omega^2
$$
$\mathrm{v}=\mathrm{R} \omega$ (for pure rolling)
$$
\mathrm{K} . E C=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} 	imes \frac{2}{3} m \mathrm{R}^2 	imes \frac{\mathrm{v}^2}{\mathrm{R}^2}=\frac{5}{6} m \mathrm{v}^2
$$
Energy remains conserve during whole journey.
$\mathrm{K}_{\cdot} \mathrm{E}_i+\mathrm{P.E}_{\cdot i}=\mathrm{K}_{\cdot} \mathrm{E}_f+\mathrm{P.E}_{\cdot f}$
$$
\begin{aligned}
& \Rightarrow \frac{5}{2} m \mathrm{v}^2=m g \mathrm{H} ~~~~~~~(\because {K.E.}_f=0)\\
& \Rightarrow \mathrm{H}=\frac{5}{6} 	imes \frac{\mathrm{v}^2}{g} \\
& =\frac{5 	imes(3)^2}{6 	imes 10} \\
& =\frac{15}{20} \mathrm{~m}=0.75 \mathrm{~m}=75 \mathrm{~cm}
\end{aligned}
$$

Centre of Mass and Rotational Motion

A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3 \mathrm{~m} / \mathrm{s}$ (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.

Choose the correct answer: