The 2.5 kg balls are attached to the ends of a thin rod of length 52.0 cm and ne

Question

The 2.5 kg balls are attached to the ends of a thin rod of length 52.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (as shown), a 59.0 g wad of wet putty drops onto one of the balls, hitting with a speed of 2.00 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? rad/s (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? KEafter / KEbefore = (c) Through what angle will the system rotate before it momentarily stops?

Explanation / Answer

Given Info:

mass of ball, M = 2.5 kg
L = 0.52 m
mass of putty, m = 0.0590 kg


a) The moment of inertia after the system after collision is
I = (2M + m)*(L/2)^2
I = 0.342 kg m^2

Applying conversation of angular momentum before and after the collision we get,
mv(L/2) = I
= mv(L/2) / I
= 0.0897 rad/s

b) the kinetic energy of the system after the collision
KE final = .5*I^2
the kinetic energy of the system before the collision
KE initial = .5*mv^2

Then the ratio is KE final over KE initial like so:

(.5*I^2) / .5*mv^2

This can be simplified all the way down to:

m / (2M + m), which is equal to:

.059 kg / 2(2.5) + .059

the ratio = K'/K = 0.01166

c) From the conservation of energy,
Ki + Ui = Kf + Uf we can derive the following:

.5*I^2 + mg(L/2) = 0 + mg(L/2)(1 - cos theta)

cos theta = .5*I^2 / mg(L/2)

= .5(.342)(.0897)^2 / (.059)(9.8)(.52/2)
cos theta = .0092

Then we set theta = cos-1 (.0092)
theta = 89.48 degrees

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