Collision involving angular and linear motion

In summary, to find the angle θ at which the rod stops after the collision, you can use the conservation of angular momentum by taking the cross product of the block's linear velocity and its distance from point O, which gives you its angular momentum. This will be conserved before and after the collision, allowing you to find the initial and final angular velocities and ultimately the angle θ.
  • #1
GwtBc
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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}})

##
 
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  • #2
GwtBc said:
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?
 
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  • #3
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.
 
  • #4
Have you covered the topic of the angular momentum of a single particle relative to a specified origin?
 
  • #5
TSny said:
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?
 
  • #6
GwtBc said:
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.
 
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1. What is the difference between angular and linear motion?

Angular motion refers to the rotation of an object around a fixed point, while linear motion refers to the movement of an object in a straight line.

2. How does angular motion affect linear motion in a collision?

In a collision, angular motion can transfer to linear motion and vice versa. This means that an object rotating with angular velocity can cause another object to move in a linear direction.

3. Can angular and linear motion occur at the same time?

Yes, it is possible for an object to have both angular and linear motion at the same time. This is seen in cases where an object is both rotating and moving in a straight line.

4. What factors determine the outcome of a collision involving angular and linear motion?

The outcome of a collision involving angular and linear motion depends on various factors such as the mass and velocity of the objects, the angle of collision, and the point of impact.

5. How is momentum conserved in a collision involving angular and linear motion?

Momentum is conserved in a collision involving angular and linear motion by the transfer of angular and linear momentum between the colliding objects. This means that the total momentum of the system before and after the collision remains the same.

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