I do not know how to solve this problem. The 2.5 kg balls are attached to the en

Question

I do not know how to solve this problem.

The 2.5 kg balls are attached to the ends of a thin rod of length 44.0 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 74.0 g wad of wet putty drops onto one of the balls, hitting with a speed of 3.00 m/s and then sticking to it. What is the angular speed of the system just after the putty wad hits? rad/s What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? KE after / KE before = Through what angle will the system rotate before it momentarily stops? Degree

Explanation / Answer

a.)mv(d/2)=Iw

I=(2M+m)(d/2)^2

So w=2mv/((2M+m)*d)=2*0.074*3/(2*2.5+0.074)*0.44=0.1988 rad/s

b.)Kf/Ki=0.5Iw^2/0.5mv^2=(2*2.5+0.074)(0.22)^2*(0.1988)^2/0.074*9=0.0146

c.)cos(theta)=0.0304

So theta=88.26

So angle=88.26+90=178.25 degrees

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