Using conservation of mechanical energy: the sum of the rotational and gravitational potential energy when the rod is horizontal is equal to the sum when the rod is vertical.
When the rod is horizontal, there is no rotational kinetic energy, and the gravitational potential energy is given by:
PE_horizontal = m * g * h
where m = 2kg (mass of rod), g = 9.8m/s^2 (acceleration due to gravity), and h = 1.2m (distance of center of mass to the axis). Thus,
PE_horizontal = 2 * 9.8 * 1.2 = 23.52 Joulse (J).
When the rod is vertical, it has no gravitational potential energy (since the center of gravity is the axis), but it has rotational kinetic energy given by:
KE_vertical = 0.5 * I * ω^2
where I = 4.0kgm^2 (moment of inertia) and ω is the angular speed when the rod is vertical that we want to find.
Since the sum of energies is constant, we have:
23.52 = 0.5 * 4.0 * ω^2
ω^2 = 23.52/(0.5*4) = 11.76
ω = √(11.76) = 3.43 rad/s
So the angular speed when the rod is vertical is 3.43 rad/s.
A nonuniform 2.0-kg rod is 2.0 m long. The rod is mounted to rotate freely about a horizontal axis perpendicular to the rod that passes through one end of the rod. The moment of inertia of the rod about this axis is 4.0 kg m2. The center of mass of the rod is 1.2 m from the axis. If the rod is released from rest in the horizontal position, what is its angular speed as it swings through the vertical position?
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1 answer