College Physics Problem. Please show all work. A 2.60-kg cylindrical rod of leng

Question

College Physics Problem. Please show all work.

A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed v1 12.5 m/s to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle before swinging back toward its original position. What is the angular speed of the rod immediately after the collision? 121 rad/s

Explanation / Answer

I = moment of inertia of rod about the end = ML2/3 = (2.6) (2)2/3 = 3.5 kgm2

wi = 0

r = 4L/5 = 4 (2)/5 = 1.6 m

using conservation of angular momentum of sphere -rod system

Li,sphere + Li,rod = Lf,sphere + Lf,rod

m vi r + I wi = I wf + m vf r

(0.25) (12.5) (1.6) + (3.5) (0) = (3.5) wf + (0.25) (12.5) (-1.6)

wf = 2.86 rad/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---