Question

A 75 g, 30 cm long rod hangs vertically on a friction less, horizontal axel passing through the center. A 10 g ball of clay travelling horizontally at 2.5 m/s hits and stick to the very bottom tip of the rod. To what maximum angle, measured from vertical, does the rod rotates when the ball of clay is attached to the rod?

Answer 1

Angular momentum is conserved, just before the clay hits and just after; 
mv(L/2) = Iw 

I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; 
I = [(1/12)ML^2 + m(L/2)^2] 

Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; 
(1/2)Iw^2 = mgh 

Use the "w" found in the conservation of momentum above; and solve for "h" 
h = mv^2L^2/8gI 

Next, get the angle by noting it is related to "h" as; 
h = (L/2) - (L/2)Cos() 

So finally 
Cos() = 1- 2h/L = 1 - mv^2L/4gI 

m=mass of clay 
M=mass of rod 
L=length of rod 
v=velocity of clay

Answer 2

The angle of the rod will be when the ball of clay is attached to the rod

$Cos\theta=1-\dfrac{mv^2L}{4gl}$

What is angular momentum?

Angular momentum is defined as when a body rotates then angular momentum will be the product of moment of inertia and the angular velocity of the object.

Angular momentum is conserved, just before the clay hits and just after;

$mv(\dfrac{L}{2} )= Iw$

I is the combined moment of inertia of the rod,$\dfrac{ 1}{12}ML^2$, and the clay at the tip,$m(\dfrac{L}{2})^2 ;$

$I = [(\dfrac{1}{12})ML^2 + m(\dfrac{L}{2})^2]$

Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem;

$(\dfrac{1}{2})Iw^2 = mgh$

Use the "w" found in the conservation of momentum above; and solve for "h"

$h = \dfrac{mv^2L^2}{8gI }$

Next, get the angle by noting it is related to "h" as;

$h = \dfrac{L}{2} - \dfrac{L}{2}Cos(\theta)$

So finally

$Cos(\theta) = 1- \dfrac{2h}{L }= 1 - \dfrac{mv^2L}{4gI }$

Thus the angle of the rod will be when the ball of clay is attached to the rod

$Cos\theta=1-\dfrac{mv^2L}{4gl}$

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