Collision involving angular and linear motion

[GwtBc]

Homework Statement

a small 0.199 kg block slides down a frictionless surface through height h = 0.608 m and then sticks to a uniform vertical rod of mass M = 0.398 kg and length d = 2.23 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ.

point O is at the end of the rod opposite to where the block sticks.

Homework Equations

$KE = mgh$

$v = \sqrt{2gh}$

$\omega = \frac{v}{r}$

The Attempt at a Solution

[/B]
I think I know how to do this but since it's an assignment question I want to make sure my thinking is right.
This is clearly an inelastic collision, so I can't simply equate the initial potential energy of the block to the final PE of the system.

The approach I've taken is to assert that at the instant the block collides with the rod, they'll be moving in the same direction as the block was (since the block is moving horizontally and the rod is simply resting in the vertical position hanging from point O), and linear momentum is conserved until the tension in the rod causes it to change. So the initial linear velocity of the system can be found, hence it's angular velocity and hence it's KE. This energy can then be equated to the final PE. The reason I'm unsure is because if we consider the rod + block system, then the tension is a force from within the system, so how could it affect the momentum?

Here are the equations I've used:

$v_{1} = \sqrt{2gh}$

$m_{1}\sqrt{2gh} = (m_{1} + m_{2})v_{2}$

$\Rightarrow v_{2} = \frac{m_{1}\sqrt{2gh}}{m_{1} + m_{2}}$

$\omega = \frac{v_{2}}{d}$

$\Rightarrow KE = \frac{1}{2}I(\frac{m_{1}\sqrt{2gh}}{(m_{1}+m_{2})d})^2 = \frac{1}{2}(\frac{1}{3}m_{2}d^2 + m_{1}d^2)(\frac{m_{1}\sqrt{2gh}}{(m_{1}+m_{2})d})^2$

$y_{com} = \frac{\frac{1}{2}m_{2}d+m_{1}d}{m_{1} + m_{2}}$

$KE = (m_{1} + m_{2})gl_{1}$

$l_{2} = y_{com}-l_{1}$

$\Rightarrow \theta = \arccos (\frac{l_{2}}{y_{com}})$

 

[TSny]

This is clearly an inelastic collision, so I can't simply equate the initial potential energy of the block to the final PE of the system.

Right. Good!

The approach I've taken is to assert that at the instant the block collides with the rod, they'll be moving in the same direction as the block was (since the block is moving horizontally and the rod is simply resting in the vertical position hanging from point O), and linear momentum is conserved until the tension in the rod causes it to change.

Here you need to be careful. Linear momentum of the system in the horizontal direction is only conserved if there is no external horizontal force acting during the collision. However, there is a significant horizontal external force that acts at the pivot. (Try to understand why.) So, linear momentum is not conserved during the collision.

Can you think of another conservation law that you can use for the collision?

 

[GwtBc]

conservation of angular momentum but I don't know the radius of the block's rotation before the collision. If I can find the ratio between $\omega_{1}$ and $\omega_{2}$ then I can find the ratio between the kinetic energies, then find the energy after the collision and proceed from there? But that method seems to rely a bit too much on fiddling around, so I'm not sure.

 

[TSny]

Have you covered the topic of the angular momentum of a single particle relative to a specified origin?

 

[GwtBc]

Have you covered the topic of the angular momentum of a single particle relative to a specified origin?

Oh, I think I get it now. The cross product of the block's linear velocity and it's radius from O is just the length of the rod times the linear velocity of the particle. Is that right?

 

[TSny]

Oh, I think I get it now. The cross product of the block's linear velocity and it's radius from O is constant and when the collision occurs is when the two are perpendicular so I can essentially take the length of the rod to be r. Is that right?

Yes, that's right.