Question 502: Centre of Mass and Rotational Motion

Question

A solid sphere rolling without friction on a horizontal surface with a constant speed of $2 \mathrm{~m} / \mathrm{s}$, rolls up on an inclined ramp which is inclined at $30^{\circ}$. The maximum distance travelled by the sphere on the inclined ramp is (acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2}$ )

Accepted Answer

Sphere's Kinetic Energy = Translational energy + Rotational energy
$$ \begin{aligned} & =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \ & =\frac{7}{10} m v^2 \end{aligned} $$
At maximum height h ,
Total K.E. = Potential energy
$$ \begin{aligned} & \frac{7}{10} \mathrm{mv}^2=\mathrm{mgh} \ & \mathrm{~h}=\frac{7 \mathrm{v}^2}{10 \mathrm{~g}}=\frac{7 	imes 2^2}{10 	imes 10}=0.28 \mathrm{~m} \end{aligned} $$
Distance along the incline
$$ d=\frac{h}{\sin 	heta}=\frac{0.28}{\sin \left(30^{\circ}ight)}=0.56 \mathrm{~m}=56 \mathrm{~cm} $$

Answer

56 cm

Answer

25 cm

Answer

47 cm

Answer

30 cm

Centre of Mass and Rotational Motion

Choose the correct answer: