Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2596 Upvotes
Verified Answer
The correct answer is:
decreases slightly
$m g \sin \theta-F_{r}=m a$
$F_{r}=\frac{2}{5} m r^{2} \frac{a}{r^{2}}$
$\Rightarrow a=\frac{5}{7} \frac{g \sin \theta}{R-r}$
$\omega=\sqrt{\frac{5 g}{7(R-r)}}$
$\Rightarrow T=2 \pi \sqrt{\frac{7(R-r)}{5 r}}$
$F_{r}=\frac{2}{5} m r^{2} \frac{a}{r^{2}}$
$\Rightarrow a=\frac{5}{7} \frac{g \sin \theta}{R-r}$
$\omega=\sqrt{\frac{5 g}{7(R-r)}}$
$\Rightarrow T=2 \pi \sqrt{\frac{7(R-r)}{5 r}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.