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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

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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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