Q90 / 90 · KVPY 2013 (SB/SX) · Physics

A ball is rolling without slipping in a spherical shallow bowl (radius $\mathrm{R}$ ) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled, the period of oscillation Figure

  • A
    increases slightly
  • B
    is reduced by a factor of $1 / 2$
  • C
    is increased by a factor of 2
  • Correct answer
    D
    decreases slightly

Solution

$m g \sin \theta-F_{r}=m a$ $F_{r}=\frac{2}{5} m r^{2} \frac{a}{r^{2}}$ $\Rightarrow a=\frac{5}{7} \frac{g \sin \theta}{R-r}$ $\omega=\sqrt{\frac{5 g}{7(R-r)}}$ $\Rightarrow T=2 \pi \sqrt{\frac{7(R-r)}{5 r}}$

Hence, the correct answer is option: 'D'.

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