Physics - Rotation Question with Solution | TestHub

By the law of conservation of energy, the total initial kinetic energy of the rolling solid sphere is converted into potential energy at the maximum height on the inclined ramp.

The total kinetic energy of a solid sphere rolling without slipping is given by:
$K=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

For a solid sphere, the moment of inertia is $I=\frac{2}{5}m{R}^2$ and the angular velocity is $\omega =\frac{v}{R}$.

$K=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(\frac{v}{R}\right)}^2$
$K=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$

The potential energy at the maximum distance $s$ along the incline is:
$U=mgh=mg(s\sin \theta )$

Equating kinetic and potential energy:
$\frac{7}{10}m{v}^2=mgs\sin 30^{\circ }$

Substituting the given values $v=2\text{ m/s}$, $g=10{\text{ m/s}}^2$, and $\sin 30^{\circ }=\frac{1}{2}$:
$\frac{7}{10}(2{)}^2=10\times s\times \frac{1}{2}$
$\frac{28}{10}=5s$
$s=\frac{28}{50}\text{ m}=0.56\text{ m}$

Converting to centimeters:
$s=0.56\times 100\text{ cm}=56\text{ cm}$

Answer: 56 cm