Question

A solid sphere and a solid cylinder roll down without slipping along an inclined plane. If they start from rest from the top of the inclined plane, the ratio of the velocities of the solid sphere and solid cylinder when they reach the bottom of the inclined plane is:

• A. \( \sqrt{25} : \sqrt{21} \)
• B. \( \sqrt{3} : \sqrt{2} \)
• C. \( \sqrt{25} : \sqrt{14} \)
• D. \( \sqrt{15} : \sqrt{14} \)

Hint:

For rolling objects, use conservation of energy, accounting for both translational and rotational kinetic energy. The moment of inertia determines the fraction of energy that goes into rotation.

Answer 1

The Correct Option is D

We need to find the ratio of the velocities of a solid sphere and a solid cylinder rolling down an inclined plane without slipping, starting from rest.
Step 1: Apply conservation of energy.
For both objects, potential energy at the top (\( mgh \)) converts to kinetic energy at the bottom (\( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)), where \( \omega = \frac{v}{r} \).
Step 2: Moment of inertia.
Solid sphere: \( I_{\text{sphere}} = \frac{2}{5}mr^2 \).
Solid cylinder: \( I_{\text{cylinder}} = \frac{1}{2}mr^2 \).
Step 3: Energy conservation for the sphere.
\[ m_{\text{sphere}} g h = \frac{1}{2} m_{\text{sphere}} v_{\text{sphere}}^2 + \frac{1}{2} I_{\text{sphere}} \left( \frac{v_{\text{sphere}}}{r} \right)^2 \] \[ \frac{1}{2} m_{\text{sphere}} v_{\text{sphere}}^2 + \frac{1}{2} \left( \frac{2}{5} m_{\text{sphere}} r^2 \right) \left( \frac{v_{\text{sphere}}^2}{r^2} \right) = \frac{1}{2} m_{\text{sphere}} v_{\text{sphere}}^2 \left( 1 + \frac{2}{5} \right) = \frac{1}{2} m_{\text{sphere}} v_{\text{sphere}}^2 \cdot \frac{7}{5} \] \[ m_{\text{sphere}} g h = \frac{1}{2} m_{\text{sphere}} v_{\text{sphere}}^2 \cdot \frac{7}{5} \implies v_{\text{sphere}}^2 = \frac{10}{7} g h \implies v_{\text{sphere}} = \sqrt{\frac{10}{7} g h} \] Step 4: Energy conservation for the cylinder.
\[ m_{\text{cylinder}} g h = \frac{1}{2} m_{\text{cylinder}} v_{\text{cylinder}}^2 + \frac{1}{2} I_{\text{cylinder}} \left( \frac{v_{\text{cylinder}}}{r} \right)^2 \] \[ \frac{1}{2} m_{\text{cylinder}} v_{\text{cylinder}}^2 + \frac{1}{2} \left( \frac{1}{2} m_{\text{cylinder}} r^2 \right) \left( \frac{v_{\text{cylinder}}^2}{r^2} \right) = \frac{1}{2} m_{\text{cylinder}} v_{\text{cylinder}}^2 \left( 1 + \frac{1}{2} \right) = \frac{1}{2} m_{\text{cylinder}} v_{\text{cylinder}}^2 \cdot \frac{3}{2} \] \[ m_{\text{cylinder}} g h = \frac{1}{2} m_{\text{cylinder}} v_{\text{cylinder}}^2 \cdot \frac{3}{2} \implies v_{\text{cylinder}}^2 = \frac{4}{3} g h \implies v_{\text{cylinder}} = \sqrt{\frac{4}{3} g h} \] Step 5: Compute the ratio of velocities.
\[ \frac{v_{\text{sphere}}}{v_{\text{cylinder}}} = \sqrt{\frac{\frac{10}{7} g h}{\frac{4}{3} g h}} = \sqrt{\frac{10}{7} \cdot \frac{3}{4}} = \sqrt{\frac{30}{28}} = \sqrt{\frac{15}{14}} \] Thus, the ratio \( v_{\text{sphere}} : v_{\text{cylinder}} = \sqrt{15} : \sqrt{14} \). Final Answer: \[ \boxed{\sqrt{15} : \sqrt{14}} \]