A slender rod is 80.0 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 0.0500-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. What is the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.

So, at the lowest point,

∆U = M2•g•y2 + M1•g•y1

Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is -40cm(-04m) from the midpoint

∆U = M2•g•y2 + M1•g•y1

∆U=0.05•9.81•(-0.4) + 0.02•9.82•0.4

∆U=-0.1962+0.07848

∆U=-0.11772 Nm

Now, the moment of inertia of the rod is given as

I=∫r²dm

dm=2pdr

I= 2p∫r²dr

I= 2 × 0.12/0.8 ∫r²dr; from r=0 to 0.4

I=0.3 [r³/3] from r=0 to 0.4

I= 0.3 [ 0.4³/3 -0] ,from r=0 to 0.4

I=0.3 × 0.02133

I=0.0064kg/m².

calculating of inertia of the end masses.

I(1+2)=Σmr² = (m1+m2)r²

I(1+2)=(0.02+0.05)0.4²

I(1+2)=0.07×0.4²

I(1+2)=0.0112 kg/m²

Now, the Energy of the masses due to angular velocity is given as

K.E=½ (I + I(1+2))w²

K.E=½(0.0064+0.0112)w²

K.E= 0.0088w²

Using conservation of energy

The potential energy is equal to the kinetic energy of the system

K.E=P.E

0.0088w²=0.11772

Then, w²=0.11772/0.0088

w²=13.377

w=√13.377

w=3.66rad/s

Then, the relationship between linear velocity and angular velocity is given by

v=wr

v=3.66×0.4

v=1.463m/s

The required linear speed is 1.46m/s approximately