# Homework Help: Glob and energy

1. Jan 25, 2010

### zhenyazh

Hi,
i am preparing for the test and have a problem with the following question.
an image is attached

A thin stick of mass M = 4.9 kg and length L = 2.6 m is hinged at the top. A piece of clay, mass m = 1.0 kg and velocity V = 4.9 m/s hits the stick a distance x = 2.20 m from the hinge and sticks to it. What is the angular velocity of the stick immediately after the collision?

What is the ratio of the final mechanical energy to the initial mechanical energy?

if i understand correctly, the final mechanical energy is before the bar starts to move up.
so i should take into consideration only the kinetic energy.
the initial energy is of course the energy of the clay which i regard as a point body
and thus 0.5mv^2.
the final energy is the energy of bar and clay joined.
this is 0.5Iw^2+0.5mu^2 where u is the velocity of the center of mass of the joint body.
if i know that angular speed of the joint body i can use w=ur, where r is the distance
between the centre of mass and the point with respect to which i am doing my calculations. that would be the distance from the pivot that holds the bar.
Xcm=(mx+lM)/(m+M)=2.532
that gives me
0.5Iw^2+0.5mw^2*2.532^2.
but i get the wrong answer.
where am i wrong?

thanks

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2. Jan 25, 2010

### kuruman

Correct so far.
Not correct. The energy after the collision is just (1/2)Iω2 where I is the moment of inertia of the composite body about the hinge. Use conservation of angular momentum about the hinge to relate the initial velocity of the clay to ω.

This is an inelastic collision that does not conserve mechanical energy.

Last edited: Jan 25, 2010
3. Jan 25, 2010

### zhenyazh

i will try. but can u explain when the energy is a sum of two expression and when not.
that is why do we lose the second part here

4. Jan 25, 2010

### zhenyazh

well it worked!! :)
so thanks.
but why is it only what u said?

5. Jan 25, 2010

### kuruman

Consider a rigid body that rotates about a fixed axis. Think of the body consisting of N point masses. All the masses have the same angular velocity ω. The total kinetic energy of these masses is

$$K=\sum^{N}_{i=1}\frac{1}{2}m_i v^{2}_{i}$$

But

$$v_{i}=\omega r_i$$

Therefore

$$K=\sum^{N}_{i=1}\frac{1}{2}m_i \omega^2 r^{2}_{i}=\frac{1}{2} \left( \sum^{N}_{i=1}m_i r^{2}_{i} \right) \omega^2=\frac{1}{2}I \omega^2$$

That's why.

6. Jan 25, 2010

### zhenyazh

great thanks. this seems logical.
but i saw in one of my books that the overall kinetic energy
of a body is the kinetic energy of its center of mass that is 0.5mv^2
plus the component with inertia that u mentioned.

why the difference, is there some nuance i am missing?

7. Jan 25, 2010

### kuruman

Yes. You are missing that there are two ways to write the kinetic energy of a body that instantaneously rotates and translates

Way 1
$$K=\frac{1}{2}I \omega^2$$

where I is the moment of inertia about the instantaneous axis of rotation

Way 2
$$K=\frac{1}{2}I_{CM} \omega^2+\frac{1}{2}m V_{CM}^2$$

where ICM is the moment of inertia about the center of mass and VCM is the translational speed of the center of mass.

You can derive Way 2 from Way 1 by using the parallel axes theorem. These two are equivalent but you have to pay attention which one you use and what you use for moment of inertia. You cannot mix and much as you seem to have done when you originally attempted this problem.

Last edited: Jan 25, 2010
8. Jan 25, 2010

### zhenyazh

so for example for a rod with a pivot in the middle
would 1/12ml^2 be the I or the Icm?

9. Jan 25, 2010

### kuruman

Yes, and because it is pivoted in the middle, i.e. at the center of mass, VCM = 0 and the expressions in Way 1 and Way 2 become identical.

10. Jan 25, 2010

### zhenyazh

i think u misunderstood me.
what is the difference between the general inetria of a rod and Icm?

11. Jan 25, 2010

### zhenyazh

oh and of course really thanks for helping for such long time

12. Jan 25, 2010

### kuruman

OK, I see the question now, I did misunderstand you. ICM is the moment of inertia about an axis that goes through the center of mass; I is the moment of inertia about any other axis that does not go through the center of mass.

For a rod of mass m and total length L the CM is at L/2 and

ICM=(1/12)mL2

If you want to calculate the moment of inertia about one end of the rod, you have to use the parallel axes theorem. The result is

IEnd=ICM+m(L/2)2=(1/3)mL2.

13. Jan 25, 2010