Slender rod is 90.0 cm long and has mass 0.120 kg. a small 0.0200 kg sphere is welded to one end of the rod, and a small 0.0400 kg sphere is welded to the other end. the rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest.

The linear speed is 1.33 m/s.

Explanation:

Given that,

Length of rod = 90.0 cm

Mass of slender rod = 0.120 kg

Mass of small sphere = 0.0200 kg

Mass of another small sphere = 0.0400 kg

Suppose, we need to find the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

We need to calculate the the change in potential of the complete system

Using formula of change in potential

m₂ and m₃ are the masses at the rod ends.

The rod center of mass neither gains nor loses potential

Put the value into the formula

We need to calculate the moment of inertia of the rod

Using formula of moment of inertia of the rod

Put the value into the formula

We need to calculate the moment of inertia of the end masses

Using formula of moment of inertia

Put the value into the formula

We need to calculate the change in potential energy to the system kinetic energy

Using formula of kinetic energy

Put the value into the formula

We need to calculate the linear speed

Using formula of linear speed

Put the value into the formula

Hence, The linear speed is 1.33 m/s.