Question 524: Centre of Mass and Rotational Motion

Question

A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is

Accepted Answer

Total $(\mathrm{K} . \mathrm{E})_{	ext {ring }}=\mathrm{Mv}^2\quad	ext{.... (i)}$
Total (K.E) $)_{	ext {disc }}=\frac{3}{4} \mathrm{Mv}^2\quad	ext{.... (ii)}$
$$\begin{aligned}
&	ext { Dividing equation (ii) by equation (i) }\
&\begin{aligned}
& \frac{(\mathrm{K} . \mathrm{E})_{	ext {disc }}}{(\mathrm{K} . \mathrm{E})_{	ext {ring }}}=\frac{\frac{3}{4} \mathrm{Mv}^2}{\mathrm{Mv}^2} \
	herefore \quad & (\mathrm{~K} . \mathrm{E})_{	ext {disc }}=(\mathrm{K} . \mathrm{E})_{	ext {ring }} 	imes \frac{3}{4}=6 	imes \frac{3}{4}=\frac{9}{2} \mathrm{~J}
\end{aligned}
\end{aligned}$$

Answer

$\frac{3}{2} \mathrm{~J}$

Answer

$\frac{5}{2} \mathrm{~J}$

Answer

$\frac{7}{2} \mathrm{~J}$

Answer

$\frac{9}{2} \mathrm{~J}$

Centre of Mass and Rotational Motion

A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is

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