To solve this problem, we need to determine the angular velocity of the rod at the instant \(\theta = 60^\circ\). We will use the principles of work and energy to solve this. ### Given: - Length of the rod, \( L = 2 \, \text{m} \) - Mass of the rod, \( m = 30 \, \text{kg} \) - Force applied, \( F = 40 \, \text{N} \) - Initial angle, \( \theta = 0^\circ \) - Final angle, \( \theta = 60^\circ \) - Neglect friction and the masses of blocks A and B. ### Steps to solve: 1. **Determine the work done by the force \( F \):** The work done by the force \( F \) as the rod moves from \(\theta = 0^\circ\) to \(\theta = 60^\circ\) can be calculated by considering the horizontal displacement of point B. When \(\theta = 0^\circ\), point B is at the origin. When \(\theta = 60^\circ\), the horizontal displacement \( x_B \) of point B is: \[ x_B = L \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m} \] The work done by the force \( F \) is: \[ W = F \cdot x_B = 40 \, \text{N} \times \sqrt{3} \, \text{m} = 40\sqrt{3} \, \text{J} \] 2. **Determine the change in potential energy:** The center of mass of the rod moves vertically as the rod rotates. Initially, when \(\theta = 0^\circ\), the center of mass is at a height of \( L/2 \) from the bottom. When \(\theta = 60^\circ\), the vertical height \( h \) of the center of mass is: \[ h = \frac{L}{2} \cos(60^\circ) = \frac{2}{2} \times \frac{1}{2} = \frac{1}{2} \, \text{m} \] The change in potential energy \( \Delta PE \) is: \[ \Delta PE = m g \Delta h = 30 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \left( \frac{1}{2} - 1 \right) \, \text{m} = 30 \times 9.81 \times \left( -\frac{1}{2} \right) = -147.15 \, \text{J} \] 3. **Apply the work-energy principle:** The work done by the force \( F \) is equal to the change in kinetic energy plus the change in potential energy: \[ W = \Delta KE + \Delta PE \] Since the rod starts from rest, the initial kinetic energy is zero. The final kinetic energy is: \[ \Delta KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the rod about the center of mass, and \( \omega \) is the angular velocity. For a rod rotating about its center, the moment of inertia \( I \) is: \[ I = \frac{1}{12} m L^2 = \frac{1}{12} \times 30 \, \text{kg} \times (2 \, \text{m})^2 = 10 \, \text{kg} \cdot \text{m}^2 \] Therefore: \[ 40\sqrt{3} = \frac{1}{2} \times 10 \times \omega^2 - 147.15 \] Solving for \(\omega\): \[ 40\sqrt{3} + 147.15 = 5 \omega^2 \] \[ 69.28 + 147.15 = 5 \omega^2 \] \[ 216.43 = 5 \omega^2 \] \[ \omega^2 = \frac{216.43}{5} = 43.286 \] \[ \omega = \

To solve this problem, we need to determine the angular velocity of the rod at the instant \(\theta = 60^\circ\). We will use the principles of work and energy to solve this. ### Given: - Length of the rod, \( L = 2 \, \text{m} \) - Mass of the rod, \( m = 30 \, \text{kg} \) - Force applied, \( F = 40 \, \text{N} \) - Initial angle, \( \theta = 0^\circ \) - Final angle, \( \theta = 60^\circ \) - Neglect friction and the masses of blocks A and B. ### Steps to solve: 1. **Determine the work done by the force \( F \):** The work done by the force \( F \) as the rod moves from \(\theta = 0^\circ\) to \(\theta = 60^\circ\) can be calculated by considering the horizontal displacement of point B. When \(\theta = 0^\circ\), point B is at the origin. When \(\theta = 60^\circ\), the horizontal displacement \( x_B \) of point B is: \[ x_B = L \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m} \] The work done by the force \( F \) is: \[ W = F \cdot x_B = 40 \, \text{N} \times \sqrt{3} \, \text{m} = 40\sqrt{3} \, \text{J} \] 2. **Determine the change in potential energy:** The center of mass of the rod moves vertically as the rod rotates. Initially, when \(\theta = 0^\circ\), the center of mass is at a height of \( L/2 \) from the bottom. When \(\theta = 60^\circ\), the vertical height \( h \) of the center of mass is: \[ h = \frac{L}{2} \cos(60^\circ) = \frac{2}{2} \times \frac{1}{2} = \frac{1}{2} \, \text{m} \] The change in potential energy \( \Delta PE \) is: \[ \Delta PE = m g \Delta h = 30 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \left( \frac{1}{2} - 1 \right) \, \text{m} = 30 \times 9.81 \times \left( -\frac{1}{2} \right) = -147.15 \, \text{J} \] 3. **Apply the work-energy principle:** The work done by the force \( F \) is equal to the change in kinetic energy plus the change in potential energy: \[ W = \Delta KE + \Delta PE \] Since the rod starts from rest, the initial kinetic energy is zero. The final kinetic energy is: \[ \Delta KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the rod about the center of mass, and \( \omega \) is the angular velocity. For a rod rotating about its center, the moment of inertia \( I \) is: \[ I = \frac{1}{12} m L^2 = \frac{1}{12} \times 30 \, \text{kg} \times (2 \, \text{m})^2 = 10 \, \text{kg} \cdot \text{m}^2 \] Therefore: \[ 40\sqrt{3} = \frac{1}{2} \times 10 \times \omega^2 - 147.15 \] Solving for \(\omega\): \[ 40\sqrt{3} + 147.15 = 5 \omega^2 \] \[ 69.28 + 147.15 = 5 \omega^2 \] \[ 216.43 = 5 \omega^2 \] \[ \omega^2 = \frac{216.43}{5} = 43.286 \] \[ \omega = \

A uniform rod of mass m and length  can rotate freely about a horizontal axis passing through end A. If it is released from an angle of 60° with vertical as shown, then find the speed of end B when it passes through lower most position.

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A pendulum consists of a 4.00 kg mass suspended at the end of a 75.0 cm thin, massless rod. The pendulum is released from an angle of 20.0 degrees with the vertical. What is the velocity of the pendulum as it passes through the origin?

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