Question 479: Centre of Mass and Rotational Motion

Question

A solid sphere at rests rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of plane is ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

Step 1: Energy conservation
At the top, the sphere has potential energy:
$ E_{	ext{top}} = mgh $
At the bottom, the energy is entirely kinetic (translational + rotational):
$ E_{	ext{bottom}} = 	frac{1}{2} m v^2 + 	frac{1}{2} I \omega^2 $
Step 2: Moment of inertia of a solid sphere
For a solid sphere about its diameter:
$ I = 	frac{2}{5} mR^2 $
Also, rolling without slipping condition:
$ v = \omega R \quad \Rightarrow \quad \omega = 	frac{v}{R} $
Step 3: Substitute
$ E_{	ext{bottom}} = 	frac{1}{2} m v^2 + 	frac{1}{2}\left(	frac{2}{5} mR^2ight)\left(	frac{v}{R}ight)^2 $
$ = 	frac{1}{2} m v^2 + 	frac{1}{5} m v^2 = 	frac{7}{10} m v^2 $
Step 4: Equating to initial energy
$ m g h = 	frac{7}{10} m v^2 $
Cancel $m$:
$ g h = 	frac{7}{10} v^2 $
$ v^2 = 	frac{10}{7} g h $
$ v = \sqrt{	frac{10 g h}{7}} $
Ã¢ÂÂ Final Answer:
The speed of the sphere at the bottom is:
$ \boxed{\left(\frac{10gh}{7}ight)^{1/2}} $
So the correct option is
B
.

Answer

$\left(\frac{9 \mathrm{gh}}{11}\right)^{\frac{1}{2}}$

Answer

$\left(\frac{10 g h}{7}\right)^{\frac{1}{2}}$

Answer

$\left(\frac{8 \mathrm{gh}}{7}\right)^{\frac{1}{2}}$

Answer

$\left(\frac{6 \mathrm{gh}}{7}\right)^{\frac{1}{2}}$

Centre of Mass and Rotational Motion

A solid sphere at rests rolls down an inclined plane of vertical height h without sliding. Its speed on reaching the bottom of plane is ( \mathrm{g}= acceleration due to gravity)

Choose the correct answer: