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## Homework Statement

Figure: http://i.imgur.com/E2D1hkW.png

A massless rod of length 2[itex]R[/itex] is attached at the middle to a pivot point that allows it to rotate in the vertical plane. Masses [itex]m[/itex] and 2[itex]m[/itex] are attached to the rod at the locations depicted in the figure. Initially the rod makes an angle of [itex]\theta[/itex] with the vertical. A lump of clay, with a mass of 3[itex]m[/itex] and a velocity [itex]v[/itex], strikes the end of the rod opposite to where mass [itex]m[/itex] is attached with and sticks to the rod.

a. What is the magnitude of the angular velocity ω for the rod just after it is struck by the clay?

b. What is the magnitude and direction of the net torque [itex]\Gamma[/itex] on the rod just after the clay mass sticks to the rod

c. Does the kinetic energy of the system change when the clay mass impacts the rod? If so, does it increase or decrease and by how much?

## Homework Equations

[itex]\Gamma[/itex] = I [itex]\frac{dω}{dt}[/itex]

[itex]\Gamma[/itex] =

**r**x

**F**

E

_{rotational}= [itex]\frac{1}{2}[/itex]Iω

^{2}

Angular Momentum= L =

**r**x

**p**, where p is the linear momentum.

L = Iω

## The Attempt at a Solution

[itex]For a:[/itex] There is no external torque on the system if we think of the system as the clay and rod so we can use conservation of momentum.

Initially, before the clay hits the rod, the rod has no angular momentum but the clay does and is given by:

L

_{before collision}=

**r**x

**p**= [itex]rvmsin\alpha[/itex] where [itex]\alpha[/itex] is the angle between [itex]r[/itex] and [itex]v[/itex] (Figure for this: http://i.imgur.com/GY9E1hZ.png )

But since sin[itex]\alpha[/itex] = [itex]\frac{Rsin\theta}{r}[/itex], replacing this in the previous equation we get

L

_{before collision}= [itex]vmRsin\theta[/itex]

After the clay hits the rod, the rod and clay combo have an angular momentum given by:

L

_{after collision}= Iω , where I is the moment of inertia.

Finding the moment of inertia:

I = mR

^{2}+ 2m([itex]\frac{1}{2}[/itex] R)

^{2}+ 3mR

^{2}

= 4mR

^{2}+ ([itex]\frac{1}{2}[/itex]mR)

^{2}

=[itex]\frac{9}{2}[/itex] mR

^{2}

By the conservation of angular momentum we have:

L

_{before collision}= L

_{after collision}

vmRsin[itex]\theta[/itex] = [itex]\frac{9}{2}[/itex] mR

^{2}ω

Solving for ω we get

ω = [itex]\frac{2}{9}[/itex][itex]\frac{vsin\theta}{R}[/itex]

However, since the net force on the system (the clay and rod again) is 0 we should be able to get ω through the conservation of energy

E

_{before collision}= E

_{after collision}

[itex]\frac{1}{2}[/itex]mv

^{2}= [itex]\frac{1}{2}[/itex]Iω

^{2}

But then we get a completely different ω:

ω = [itex]\frac{v}{R}[/itex] [itex]\sqrt{\frac{2}{9}}[/itex]

Clearly one of them is incorrect and it seems to be the second equation for ω since it does not depend on the angle[itex]\theta[/itex]. If [itex]\theta[/itex] was 0 I would imagine the torque to be larger and thus ω be larger

[itex]For b[/itex]: I'm having an even harder time with this one. My train of thought was this: since we don't know the Force we can't use [itex]\Gamma[/itex] =

**r**x

**F**, we also can't use [itex]\Gamma[/itex] = I [itex]\frac{dω}{dt}[/itex] since we don't have an equation for how ω changes during the impact. After the collision, ω is constant because there is no net torque. So that leaves us with the energy equation E

_{rotational}= [itex]\frac{1}{2}[/itex]Iω

^{2}but how would we relate this to torque?

[itex]For c[/itex]: Since there is no net force on the system (clay+rod) energy is conserved and the only energy associated is kinetic energy and so kinetic energy is conserved.

But from question a, we used this assumption and got a different (incorrect?) ω.

Any help or suggestions would be greatly appreciated.

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