Question

A pendulum of mass \( 1 \) kg and length \( l = 1 \) m is released from rest at an angle \( \theta = 60^\circ \). The power delivered by all the forces acting on the bob at angle \( \theta = 30^\circ \) will be (Take, \( g = 10 \) m/s\( ^2 \)):

• A. \( 13.4 \) W
• B. \( 20.4 \) W
• C. \( 24.6 \) W
• D. zero

Hint:

Power is given by \( P = F v \cos \theta \). For pendulum motion, velocity can be found using energy conservation.

Answer 1

The Correct Option is A

A pendulum of mass \( m = 1 \) kg and length \( l = 1 \) m is released from rest at an angle \( \theta_0 = 60^\circ \). We need to find the power delivered by all forces acting on the bob when the angle \( \theta = 30^\circ \). The forces acting on the bob are gravity and tension. The power at any instant is given by \( P = F \cdot v \cdot \cos(\phi) \), where \( F \) is the force, \( v \) is the velocity, and \(\phi\) is the angle between the force and velocity vector. For this scenario, the power is delivered by the component of gravitational force along the direction of motion.

The velocity of the pendulum bob at any point can be determined using energy conservation. Initially, the gravitational potential energy (GPE) at the starting point \(\theta_0\) is converted into kinetic energy (KE) and GPE at \(\theta=30^\circ\).

Step 1: Initial and Final GPE
The initial height \( h_0 \) when \(\theta_0=60^\circ\) from the vertical is given by \( h_0=l-l\cos(\theta_0)=1-\cos(60^\circ)=1-\frac{1}{2}=\frac{1}{2} \text{ m}\). The initial GPE is \( U_0=mgh_0=1 \times 10 \times \frac{1}{2}=5 \text{ J}\). The height at \(\theta=30^\circ\) is \( h_1=l-l\cos(30^\circ)=1-\cos(30^\circ)=1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2} \text{ m}\). The GPE at \(\theta=30^\circ\) is \( U_1=mgh_1=1 \times 10 \times \frac{2-\sqrt{3}}{2} \).

Step 2: Conservation of Energy to find velocity at \(\theta=30^\circ\)
The initial kinetic energy is zero since it starts from rest. At \(\theta=30^\circ\), using conservation of energy, \( U_0=U_1+KE \).
\( 5=\frac{2-\sqrt{3}}{2} \times 10 + \frac{1}{2}mv^2 \)
Simplifying gives:
\( 5=10(\frac{2-\sqrt{3}}{2})+\frac{v^2}{2} \)
\( \frac{v^2}{2}=5-10(\frac{2-\sqrt{3}}{2}) \)
\( v^2=2[5-5(2-\sqrt{3})]\)
\( v^2=[10-10+5\sqrt{3}]=5\sqrt{3} \)
\( v=\sqrt{5\sqrt{3}} \)

Step 3: Calculation of Power
At angle \(\theta\), only the tangential component of gravitational force does work: \( F_{\text{tangent}}=mg\sin\theta \). Therefore:
\( F_{\text{tangent}}=1 \times 10 \times \sin(30^\circ)=10 \times \frac{1}{2}=5 \text{ N}\)
The power \( P=F_{\text{tangent}} \cdot v \). Plugging in the values, \( P=5 \text{ N} \cdot \sqrt{5\sqrt{3}} \text{ m/s} \)
\( P=5\sqrt{5\sqrt{3}} \approx 13.4 \text{ W} \).

Therefore, the power delivered by the forces acting on the pendulum at \( \theta=30^\circ \) is \( 13.4 \text{ W} \).

Answer 2

Step 1: {Find velocity at \( \theta = 30^\circ \)}
Using energy conservation: \[ v = \sqrt{2 g h} \] Step 2: {Calculate height difference}
\[ h = l (\cos 30^\circ - \cos 60^\circ) \] \[ = 1 \times \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = 0.36 { m} \] Step 3: {Find velocity}
\[ v = \sqrt{2 \times 10 \times 0.36} \] \[ = \sqrt{7.2} = 2.68 { m/s} \] Step 4: {Find power}
\[ P = (mg v) \cos 60^\circ \] \[ = (1 \times 10 \times 2.68) \times \frac{1}{2} \] \[ = 13.4 { W} \] Thus, the correct answer is (A) 13.4 W.