# Linear and rotational motion

## Homework Statement

In this given pole, mass is "M" and length is "L" and it is on a frictionless surface as the picture describes. "G" is the center of gravity and "I" is the inertia.
A ball with a mass of "m" comes as the picture and hits the pole with a velocity of "v" and turns the opposite side and leaves at the same velocity. Then the pole moves under both rotational and linear motions.

## Homework Equations

(a)
i. Write an equation for the momentum of the ball before the collision.
ii. Considering only the linear motion of the pole, write an equation for the velocity of the pole "V".

(b) Now consider the rotational motion.
i. If the ball hits x distance from point "G", write an equation for the angular momentum of the ball around point "G".
ii. Write an equation for the angular velocity of the pole around point "G"

## The Attempt at a Solution

(a)
i. P = mv
ii. → mv + M0 = -mv + MV
∴ V = $\frac{2mv}{M}$

(b)
i. L = mvx
ii. → mvx + 0 = -mvx + Iω
∴ ω = $\frac{2mvx}{I}$

#### Attachments

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tiny-tim
Homework Helper
Hi Knightycloud!

(btw, we'd say "opposite way" rather than "opposite side" )

Yes, that's fine …

what is worrying you about that?​

Yeah!!! :D
well did I get that b part correct???

tiny-tim
Homework Helper
yes

Oh, So did i get that angular velocity correct???
then the question asks at a certain value of x (lets say it's y) the point A becomes still. How to calculate that value Y???

tiny-tim
Homework Helper
let's see … A is at distance -L/2 from the centre

ok, from (a) and (b), what is the initial velocity of A?

Velocity of point A is mvxL/I
and thank you, I found the solution to the rest of the question :D I've been a bit lossing my self confidence!!! You know even if I had the answer, I believed it wrong!!!

That Y distance is = 2I/ML by the way!!!
Thanks Tim!!!