# Rigid rotational dynamics problem [closed]

A thin uniform rod AB of mass $M$ and length $L$ is free to rotate in a vertical plane about a horizontal axle at end A. A piece of putty, also of mass $M$, is thrown with velocity $V$ horizontally at the lower end B while the bar is at rest. The putty sticks to the bar. What is the minimum velocity of the putty before impact that will make the bar rotate all the way around A?

(Gravity acts on the rod)

My strategy in solving this problem consisted of trying to find the kinetic energy of the system immediately after collision and then using the work done by gravity to find the minimum velocity.

$$L = m v l$$

The centre of mass of the system must then be $3L\over{4}$ the way down from the pivot A. Hence, the moment of inertia of the centre of mass about the pivot can be derived as:

$$I_{cm} = 2m ({3l\over{4}})^2$$

From which the angular velocity may be determined by the conservation of angular momentum:

$$w = {L \over{ I_{cm} }} = {m v l \over{2m({3l\over{4}})^2}} = {8v \over{9l}}$$

Hence, the final kinetic energy $$KE = 0.5 * 2m ({v\over{2}})^2 + 0.5 (2m({3l\over{4}})^2) * ({8v\over{9l}})^2$$

However, this is incorrect since the final kinetic energy is greater than the initial kinetic energy. The final answer should be $v = \sqrt{8gL}$, according to the book, however my attempts have been in vain. How should I go about this problem?

• Your moment of inertia is wrong. – Ben51 Jan 24 '18 at 20:40
• Possible duplicate of Off-center impulse equations – ja72 Jan 24 '18 at 20:56
• For the general 3D impact problem between rigid bodies see this answer. – ja72 Jan 24 '18 at 20:57