SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 81

88. A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is

[1994]

(a) $2 / 5$

(b) $2 / 7$

(d) $3 / 7$

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Answer:

Correct Answer: 88. (b)

Solution:

(b) $\frac{K_{r}}{E} = \frac{\frac{1}{2} M K^{2} ω^{2}}{\frac{1}{2} M ω^{2} [K^{2} + R^{2}]} = \frac{K^{2}}{K^{2} + R^{2}}$

$= \frac{2 / 5}{1 + 2 / 5} = \frac{2}{7}$

Here, $K^{2} = \frac{2}{5} R^{2}$

When a body is executing only linear motion, its K.E. is given by $E_{t} = \frac{1}{2} m v^{2}$ when a body is rolling without slipping, its centre of mas has linear motion too.

Therefore, its total kinetic energy

$E = E_{t} + E_{r}$

$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

$= \frac{1}{2} m v^{2} + \frac{1}{2} m (K^{2}) \frac{v^{2}}{R^{2}}$

$= \frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})$

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System of Particles and Rotational Motion - Result Question 81

88. A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is

Answer:

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