Question 634: Centre of Mass and Rotational Motion

Question

A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (Kt) as well as rotational kinetic energy
(Kr) simultaneously. The ratio Kt
: (Kt + Kr)
for the sphere is

Accepted Answer

In rolling motion,
Translational kinetic energy, K
t
= $${1 \over 2}m{v^2}$$
Rotational kinetic energy, K
r
= $${1 \over 2}I{\omega ^2}$$
$$ 	herefore $$ K
t
+ K
r
= $${1 \over 2}m{v^2}$$ + $${1 \over 2}I{\omega ^2}$$
= $${1 \over 2}m{v^2} + {1 \over 2}\left( {{2 \over 5}m{r^2}} ight){\left( {{v \over r}} ight)^2}$$
= $${7 \over {10}}m{v^2}$$
$$ 	herefore $$ $${{{K_t}} \over {{K_t} + {K_r}}} = {{{1 \over 2}m{v^2}} \over {{7 \over {10}}m{v^2}}}$$ = $${5 \over 7}$$

Answer

7 : 10

Answer

5 : 7

Answer

10 : 7

Answer

2 : 5

Centre of Mass and Rotational Motion

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is

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