Question

A thin rod AB of length L is hinged at A such that it is free to rotate parallel to the inclined plane. Its linear mass density varies as $\frac{dm}{dx}=kx$ where x is distance from end A and k is a constant. The time period of small oscillations if it is slightly displaced from equilibrium position is $T = 2\pi \sqrt{\frac{NL}{10g}}$ where $N = \_\_\_\_\_$.

Answer 1

15

Answer 2

Solution:

1. Center of Mass (CM):
   The linear mass density is
   $\dfrac{dm}{dx}= kx$.
   Total mass:
   $M=\int_0^L kx\,dx=\frac{kL^2}{2}$.
   CM location:
   $x_{cm}=\frac{1}{M}\int_0^L x\,dm=\frac{\int_0^L x(kx)dx}{\frac{kL^2}{2}}=\frac{kL^3/3}{kL^2/2}=\frac{2L}{3}$.

2. Moment of Inertia about A:
   $I=\int_0^L x^2\,dm=\int_0^L x^2(kx\,dx)=k\int_0^L x^3dx=\frac{kL^4}{4}$.
   Thus,
   $\frac{I}{M}=\frac{\frac{kL^4}{4}}{\frac{kL^2}{2}}=\frac{L^2}{2}$.

3. Effective Gravitational Acceleration:
   Since the rod is free to rotate in a plane parallel to the inclined plane (inclination $\alpha=30^\circ$) the restoring force is due to the component of gravity along the plane. Thus, use
   $g_{\text{eff}}=g\sin30^\circ=\frac{g}{2}$.

4. Using the Physical Pendulum Formula:
   For small oscillations the period is given by

   $$T = 2\pi\sqrt{\frac{I}{M\,g_{\text{eff}}\,x_{cm}}}.$$

   Substitute $I/M=\frac{L^2}{2}$, $x_{cm}=\frac{2L}{3}$ and $g_{\text{eff}}=\frac{g}{2}$:

   $$T = 2\pi\sqrt{\frac{L^2/2}{(g/2)(2L/3)}} = 2\pi\sqrt{\frac{L^2/2}{\frac{gL}{3}}} = 2\pi\sqrt{\frac{3L}{2g}}.$$

5. Comparing with Given Expression:
   Given that

   $$T =2\pi \sqrt{\frac{NL}{10\,g}},$$

   equate the expressions under the square root:

   $$\frac{3L}{2g}=\frac{NL}{10g}.$$

   Cancel $L/g$ to obtain:

   $$\frac{3}{2}=\frac{N}{10}\quad\Longrightarrow\quad N=10\cdot\frac{3}{2}=15.$$

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Explanation (Minimal):

• Compute total mass $M=\frac{kL^2}{2}$ and $x_{cm}=\frac{2L}{3}$.
• Find moment of inertia about A: $I=\frac{kL^4}{4}$.
• Use physical pendulum formula with effective $g_{\text{eff}}=g\sin30 =\frac{g}{2}$:
  $T=2\pi\sqrt{\frac{I}{M\,g_{\text{eff}}\,x_{cm}}}=\displaystyle2\pi\sqrt{\frac{3L}{2g}}$.
• Equate with $T=2\pi\sqrt{\frac{NL}{10g}}$ to get $N=15$.

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Answer:
$N = 15$.