Question

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{\mathrm{x}}{5}}$ where $\mathrm{x}=$ _______.

Hint:

Use mechanical energy conservation to find the velocities of the ring and the solid sphere.

Answer 1

Correct Answer: 3 - 4

1. Mechanical energy conservation: \[ K_i + U_i = K_f + U_f \] \[ 0 + Mgh = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{R^2}\right) + 0 \] \[ v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} \]
2. Ratio of velocities: \[ \frac{v_{\text{ring}}}{v_{\text{solid sphere}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} \] \[ x = 3.5 \approx 4 \] Therefore, the correct answer is (4) 4.

Answer 2

We are given a circular ring and a solid sphere, both having the same radius \( R \), rolling down an inclined plane from rest without slipping. We need to find the ratio of their linear velocities at the bottom of the incline, expressed as:

\[ \frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{x}{5}} \] and determine the value of \( x \).

 

Concept Used:

When a rigid body rolls down an incline without slipping, the loss of potential energy is converted into both translational and rotational kinetic energies:

\[ mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]

Using the rolling condition \( v = R\omega \), we can rewrite the equation as:

\[ mgh = \frac{1}{2} m v^2 \left(1 + \frac{I}{mR^2}\right) \]

Thus, the velocity of the object at the bottom is given by:

\[ v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}} \]

Step-by-Step Solution:

Step 1: For the circular ring:

\[ I_{\text{ring}} = mR^2 \] \[ v_{\text{ring}} = \sqrt{\frac{2gh}{1 + \frac{I_{\text{ring}}}{mR^2}}} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh} \]

Step 2: For the solid sphere:

\[ I_{\text{sphere}} = \frac{2}{5}mR^2 \] \[ v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{2gh}{\frac{7}{5}}} = \sqrt{\frac{10gh}{7}} \]

Step 3: Find the ratio of their velocities.

\[ \frac{v_{\text{ring}}}{v_{\text{sphere}}} = \frac{\sqrt{gh}}{\sqrt{\frac{10gh}{7}}} = \sqrt{\frac{7}{10}} \]

Step 4: Compare with the given form.

\[ \sqrt{\frac{7}{10}} = \sqrt{\frac{x}{5}} \]

Step 5: Equate the two expressions inside the square roots:

\[ \frac{7}{10} = \frac{x}{5} \]

Step 6: Solve for \( x \).

\[ x = \frac{7}{10} \times 5 = 3.5 \]

Final Computation & Result:

Therefore, the value of \( x \) is:

\[ \boxed{x = 3.5} \]

Final Answer: \( x = 3.5 \)