Question

A thin, cylindrical rod ℓ = 27.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.

I am looking for the; How does it compare with the speed had the ball fallen freely through the same distance of 32.0 cm?

As well as the V swing and the V fall. (There is the exact question of this on chegg already posted if needing to see the other parts of the question in order to get this answer)

Answer 1

Given is:-

Length of the rod l=27cm

mass m=1.20 kg

diameter of the ball d=10cm and mass of the ball M=2kg

Now,

Approach:- First of all we have to find the angular speed by energy conservation then we calculate the linear speed of the center of mass of the ball and finally we find free fall speed and then compare them.

So

Gain in rotataional kinetic energy = loss in gravitational potential energy

by putting all values

eq-1

Now the moment of inertial of whole system is

by putting all the values in above equation, we get

by putting this value in eq-1 we get

Now the velocity of the center of mass

by putting the values

Note:- From here on the comparison part you are looking is being explained

Now the loss in G.P.E = gain in K.E.

thus

by plugging all the values

it's comparison with the speed had the ball fallen freely through the same distance of 32.0 cm

thus