A long, uniform rod with length L, mass M, and moment of inertia (with respect t

Question

A long, uniform rod with length L, mass M, and moment of inertia (with respect to its center of mass) Icm=(1/12)ML^2, is hinged at one end so that it can rotate, without friction, around a horizontal axis. The rod is hanging, motionless, in a vertical position when a piece of putty with same mass M is launched horizontally toward the rod and hits it at its center, with horizontal speed Vo, getting stuck to it. Write your results in terms of L, M, and Vo. (a)find the moment of inertia with respect to the hinge of the rod+putty system after the collision. (b)find the angular speed w of the rod
+putty right after the collision. (c) as the system rotates around the hinge, what is the maximum angle (theta)max it reaches before it falls back down? Assume the speed is small so that it does not do full loop.

Explanation / Answer

a)

MOI of the rod about hinged point =I = (1/12)ML^2 + M*(L/2)^2 = M*L^2/3 (parallel axis theorem)

Total MOI of ROD + PUTTY = IT = M*L^2/3 + M*(L/2)^2 = (7/12)M.L2....Ans

b)

Conservation of angular moment about the hinge:

M*V*(L/2) = IT*

=6V/7L

c)

Work done by gravity = change in rotational kinetic energy

(2M)*(L/2)*(1-cos) = (1/2)*IT*2

substituting values we get,

= cos-1(1-(3V2/14L))

Hope this helps!!

CHEERS!!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.