Question

A solid sphere of mass \(M\) and radius \(R\) is attached to a spring of negligible mass kept on a horizontal plane such that it can roll without slipping. The sphere is made to execute SHM by stretching through a distance and released. Find the time period of such oscillation. (K = spring constant)

• A. \(2\pi \sqrt{\frac{3M}{2K}}\)
• B. \(2\pi \sqrt{\frac{5K}{7M}}\)
• C. \(2\pi \sqrt{\frac{7M}{5K}}\)
• D. \(2\pi \sqrt{\frac{3K}{2M}}\)

Hint:

For rolling objects in SHM, effective mass includes rotational inertia: \(M_{\text{eff}} = M + I/R^2\). Time period: \(T = 2\pi \sqrt{M_{\text{eff}}/K}\).

Answer 1

The Correct Option is C

Step 1: Identify the system.
The solid sphere rolls without slipping on a horizontal surface, attached to a spring. The spring provides the restoring force, and the sphere's motion involves both translational and rotational kinetic energy.

Step 2: Total kinetic energy.
For a rolling solid sphere:
\[ T = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \]
where \(I = \frac{2}{5} M R^2\) for solid sphere and \(\omega = v/R\). Substituting:
\[ T = \frac{1}{2} M v^2 + \frac{1}{2} \cdot \frac{2}{5} M R^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2 \]

Step 3: Equation of motion.
For SHM, effective mass is \(M_{\text{eff}} = \frac{7}{5} M\) because of rotational motion. Spring provides force:
\[ F = -k x = M_{\text{eff}} a = \frac{7}{5} M \frac{d^2 x}{dt^2} \]

Step 4: Time period formula.
For SHM:
\[ T = 2\pi \sqrt{\frac{M_{\text{eff}}}{K}} = 2\pi \sqrt{\frac{7M/5}{K}} = 2\pi \sqrt{\frac{7M}{5K}} \]

Step 5: Verification.
Dimensionally consistent: \([T] = \sqrt{M/K}\). Accounts for both translational and rotational inertia.

Step 6: Final conclusion.
Hence, the time period of oscillation is:
\[ \boxed{2\pi \sqrt{\frac{7M}{5K}}} \]