Question

A uniform solid sphere of radius $r=\frac{R}{5}$ is placed on the inside surface of a hemispherical bowl with radius$R(=5r)$ The sphere is released from rest at an angle $\theta ={37}^{\circ }$ to the vertical and rolls without slipping. The angular speed of the sphere when it reaches the bottom of the bowl is

• A. $\sqrt{\frac{40}{7}\frac{g}{R}}$
• B. $\sqrt{\frac{20}{7}\frac{g}{R}}$
• C. $\sqrt{\frac{10}{7}\frac{g}{R}}$
• D. $\sqrt{\frac{25}{7}\frac{g}{R}}$

Answer 1

The correct option is A $\sqrt{\frac{40}{7}\frac{g}{R}}$

Draw a diagram of given problem.

Calculate initial and final energy.

Formula used:$\Delta PE+{\left(\Delta KE\right)}_{\tau }+{\left(\Delta KE\right)}_R=0$

Moment of inertia of spherical ball $=\frac{2}{5}m{r}^2$
Taking centre of the bowl as origin.

Initial potential energy $=-mg(R-r)\cos \theta$

Final potential energy $=-mg(R-r)$

Now,
Translational kinetic energy

Initial energy $=0$
Final energy $=\frac{1}{2}m{v}^2$

Then,
Rotational kinetic energy

Initial energy $=0$

Final energy $=\frac{1}{2}I{\omega }^2$

Find angular speed.
From energy conservation law,

$\Delta PE+{\left(\Delta KE\right)}_r+{\left(\Delta KE\right)}_R=0$

$-mg(R-r)(1-\cos \theta )+\frac{1}{2}m{v}_f^2+\frac{1}{2}I{\omega }^2=0$

$-mg(R-r)(1-\cos \theta )+\frac{1}{2}m.{\omega }^2{r}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2=0$

$\frac{7}{10}m{r}^2{\omega }^2=mg(R-r)(1-\cos \theta )$

${\omega }^2=\frac{10}{7r^2}g(R-r)(1-\cos \theta )$

Hence, required angular speed of ball when it reaches the ground.

$\omega =\sqrt{\frac{10g(R-r)(1-\cos \theta )}{7r^2}}$

$\omega =\sqrt{\frac{10g(R-\frac{R}{5})(1-\cos {37}^{\circ })}{7{\left(\frac{R}{5}\right)}^2}}=\sqrt{\frac{40}{7}\frac{g}{R}}$

Final answer: (b)