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Question

(A) The system acts as a compound pendulum. The time period of a compound pendulum is given by $T = 2\pi \sqrt{\frac{I_0}{M_{total} g d}}$,where $I_0$

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{2(M + 3m)\,L}}{{3(M + 2m)\,g}}} $

Vedclass · Answer

(A) The system acts as a compound pendulum. The time period of a compound pendulum is given by $T = 2\pi \sqrt{\frac{I_0}{M_{total} g d}}$,where $I_0$ is the moment of inertia about the point of suspension,$M_{total}$ is the total mass,and $d$ is the distance of the center of mass from the point of suspension.$1$. Moment of inertia about the point of suspension $O$:$I_0 = I_{rod} + I_{bob} = \frac{ML^2}{3} + mL^2 = \frac{(M + 3m)L^2}{3}$$2$. Total mass of the system:$M_{total} = M + m$$3$. Distance of the center of mass $(d)$ from the point of suspension:$d = \frac{M(L/2) + m(L)}{M + m} = \frac{(M/2 + m)L}{M + m} = \frac{(M + 2m)L}{2(M + m)}$$4$. Substituting these values into the time period formula:$T = 2\pi \sqrt{\frac{(M + 3m)L^2 / 3}{(M + m)g \cdot \frac{(M + 2m)L}{2(M + m)}}}$$T = 2\pi \sqrt{\frac{(M + 3m)L^2}{3} \cdot \frac{2(M + m)}{(M + m)g(M + 2m)L}}$$T = 2\pi \sqrt{\frac{2(M + 3m)L}{3(M + 2m)g}}$

Column-$I$	Column-$II$
$(1)$ Perpendicular Axis Theorem	$(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem	$(b)$ $I_z = I_x + I_y$

The string of a simple pendulum is replaced by a uniform rod of length $L$ and mass $M$. If the mass of the bob of the pendulum is $m$,then for small oscillations its time period would be (assume radius of bob $r << L$):

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Match Column-$I$ with Column-$II$:
Column-$I$ Column-$II$
$(1)$ Perpendicular Axis Theorem $(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem $(b)$ $I_z = I_x + I_y$

Where, $d =$ distance between two parallel axes.

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