A 2.30-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]

Verified step by step guidance

1. Identify the system and the principle to use: The pole is a rigid body rotating about its lower end, which is fixed. The principle of conservation of mechanical energy applies because no external work is done on the system.
2. Write the expression for the total mechanical energy at the initial position: The pole starts from rest, so its initial kinetic energy is zero. The potential energy is due to the center of mass of the pole, which is located at a height of \( \frac{L}{2} \) (half the length of the pole). The initial potential energy is \( U_i = m g \frac{L}{2} \), where \( m \) is the mass of the pole, \( g \) is the acceleration due to gravity, and \( L \) is the length of the pole.
3. Write the expression for the total mechanical energy just before the pole hits the ground: At this point, the potential energy is zero because the center of mass is at ground level. The pole has rotational kinetic energy given by \( K_f = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia of the pole about the pivot point and \( \omega \) is the angular velocity.
4. Relate the moment of inertia to the geometry of the pole: The moment of inertia of a uniform rod about one end is \( I = \frac{1}{3} m L^2 \). Substitute this into the expression for rotational kinetic energy.
5. Apply conservation of energy: Set the initial potential energy equal to the final rotational kinetic energy, \( m g \frac{L}{2} = \frac{1}{2} \left( \frac{1}{3} m L^2 \right) \omega^2 \). Solve for \( \omega \), the angular velocity. Then, use the relationship between linear velocity and angular velocity, \( v = \omega L \), to find the speed of the upper end of the pole just before it hits the ground.

Key Concepts

Conservation of Energy

The principle of conservation of energy states that in a closed system, the total energy remains constant over time. In this scenario, the potential energy of the pole when it is upright is converted into kinetic energy as it falls. This concept is crucial for determining the speed of the upper end of the pole just before it hits the ground.

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field. For the pole, when it is vertical, its potential energy is at a maximum, calculated using the formula PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height of the center of mass. As the pole falls, this potential energy decreases while its kinetic energy increases.

Kinetic Energy

Kinetic energy is the energy of an object in motion, given by the formula KE = 1/2 mv², where m is mass and v is velocity. As the pole falls, the potential energy it loses is converted into kinetic energy, which determines the speed of the upper end just before impact. Understanding this relationship is essential for solving the problem.