# I Determining linear velocity of pendulum

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1. Sep 21, 2016

### mmcsa

Hello,

I'm trying to develop a pendulum to test protective equipment so I want to work out the length I'll need to generate a desired velocity and the necessary mass I'll need for a specific moment of inertia. I know there are multiple ways to solve for linear velocity with equating Ek and Ep being the most common. I have also tried to do this using angular acceleration, given θ ̈ = - g/l sin⁡θ however they both seem to give me different velocities, so I think I've gone wrong somewhere!

So I have 2 methods which should mathematically be the same but I can't seem to match them

Method 1: Ek = Ep
Ep=Ek
mgh = 1/2 mv^2
v=√2gh
h = L - (Lcos⁡(-θ))
v_bottom= √(2gL(1-cos⁡(θ)))
http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section5.rhtml

Method 2: equations of angular motion

angular acceleration= θ ̈ = - g/l sin⁡θ
https://en.wikipedia.org/wiki/Inverted_pendulum#Stationary_pivot_point
ω^2= ω_o^2+2θ ̈ θ
V_tangential=ωr
V_tangential=L√(ω_o^2+2(-g/L sin⁡θ )θ )

Any top tips would be greatly appreciated

2. Sep 21, 2016

### Staff: Mentor

This gives the angular acceleration when the pendulum angle is θ. As the angle varies (as the pendulum "falls"), so does the angular acceleration.

This assumes the angular acceleration is constant, which is not true for your situation.

3. Sep 21, 2016

### mmcsa

Thanks jtbell,
Is there an alternative equation of motion I can use to work out the velocity at each point given the constantly changing angular acceleration of the system? Or is it possible with method 1 to find the resultant velocity for other cases not just at the bottom of the pendulum swing? For instance I am considering using an inverted pendulum style design where the mass is released at say 179° to vertical and it impacts at horizontal 90° (Vy=Vresultant) therefore the calculated h would be ~ pendulum rod length

4. Sep 21, 2016

### Nidum

Google ' Charpy Pendulum ' and ' Charpy impact testing '

5. Sep 21, 2016

### Staff: Mentor

Yes, you can use conservation of energy for such situations: $$E_{k,final} + E_{p,final} = E_{k,initial} + E_{p,initial} \\ \frac 1 2 mv^2_{final} + mgh_{final} = \frac 1 2 mv^2_{initial} + mgh_{initial}$$ where presumably $v_{initial} = 0$. You find the two h's from the corresponding angles by using some geometry. Of course m cancels out. This assumes that the mass of the pendulum string or rod is negligible compared to the mass of the bob.