Question

a thin rod of length 0.75 m and mass 0.37 kg is suspended freely from one end. it is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 3.6 rad/s. neglect friction and air resistance. (a) find the rod's kinetic energy at its lowest position. j (b) find how far above that position the center of mass rises. m

Answer 1

The rod's kinetic energy at its lowest position is 0.3024 J. The center of mass of the rod rises to a height of 0.339 m above its lowest position.

Step-by-step explanation:

To find the kinetic energy at the rod's lowest position, we will use the kinetic energy formula for rotational motion:

KE = (1/2)Iω².

The moment of inertia, I, for a rod rotating about an end is given by (1/3)ml².

Plugging in the values, we get I = (1/3)(0.37 kg)(0.75 m)² = 0.046875 kg·m².

Therefore, the kinetic energy KE = (1/2)(0.046875 kg·m²)(3.6 rad/s)² = 0.3024 J.

To find the height the center of mass rises, we use conservation of energy.

At the lowest point, all the energy is kinetic, and at the highest point, it is potential.

Setting KE equal to the potential energy (PE = mgh), where h is the height above the lowest position,

we get mgh = (1/2)Iω².

Solving for h, h = (Iω²)/(2mg) = (0.046875 kg·m²)(3.6 rad/s)²/(2·0.37 kg·9.81 m/s²) = 0.339 m.