A pendulum, with mass \( m = 1 \) kg and length \( l = 1 \) m, is released from rest at an initial angle \( \theta_0 = 60^\circ \). The objective is to determine the power delivered by all forces acting on the bob when it reaches an angle \( \theta = 30^\circ \). The forces involved are gravity and tension. Instantaneous power is defined as \( P = F \cdot v \cdot \cos(\phi) \), where \( F \) is the force magnitude, \( v \) is the velocity magnitude, and \(\phi\) is the angle between the force and velocity vectors. In this case, power is solely contributed by the component of gravitational force acting along the direction of motion.
The velocity of the pendulum bob at any position is found using the principle of energy conservation. The initial gravitational potential energy (GPE) at the starting angle \(\theta_0\) is transformed into kinetic energy (KE) and GPE at the angle \(\theta=30^\circ\).
Step 1: Initial and Final Gravitational Potential Energy (GPE)
The initial height \( h_0 \) relative to the equilibrium position (vertical) at \(\theta_0=60^\circ\) is calculated as \( h_0=l-l\cos(\theta_0)=1-\cos(60^\circ)=1-\frac{1}{2}=\frac{1}{2} \text{ m}\). Consequently, the initial GPE is \( U_0=mgh_0=1 \times 10 \times \frac{1}{2}=5 \text{ J}\). At \( \theta=30^\circ \), the height \( h_1 \) 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 this angle is \( U_1=mgh_1=1 \times 10 \times \frac{2-\sqrt{3}}{2} \text{ J}\).
Step 2: Energy Conservation to Determine Velocity at \(\theta=30^\circ\)
The initial kinetic energy is zero as the bob starts from rest. By applying conservation of energy at \( \theta=30^\circ \), we have \( U_0=U_1+KE \).
\( 5 \text{ J} = \left( 1 \times 10 \times \frac{2-\sqrt{3}}{2} \right) \text{ J} + \frac{1}{2}mv^2 \)
Substituting \( m=1 \) kg and simplifying:
\( 5 = 5(2-\sqrt{3}) + \frac{v^2}{2} \)
\( 5 = 10-5\sqrt{3} + \frac{v^2}{2} \)
\( \frac{v^2}{2} = 5 - (10-5\sqrt{3}) = 5\sqrt{3}-5 \)
\( v^2 = 10\sqrt{3}-10 \)
\( v = \sqrt{10\sqrt{3}-10} \text{ m/s} \)
Step 3: Power Calculation
At any angle \( \theta \), the tangential component of the gravitational force is \( F_{\text{tangent}}=mg\sin\theta \). This component is responsible for the work done and thus the power delivered. At \( \theta=30^\circ \):
\( F_{\text{tangent}} = 1 \times 10 \times \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \text{ N} \)
The power \( P \) is then calculated as \( P=F_{\text{tangent}} \cdot v \). Substituting the values:
\( P = 5 \text{ N} \cdot \sqrt{10\sqrt{3}-10} \text{ m/s} \)
\( P \approx 13.4 \text{ W} \).
Therefore, the power delivered by the forces acting on the pendulum bob at \( \theta=30^\circ \) is approximately \( 13.4 \text{ W} \).