 Posers and Puzzles

1. 24 Oct '11 10:33
A weight has been attached by a taut piece of string to point O, and is currently at rest at point A1, directly to the left from O. As it is released it goes along the semicircular path, through A2 down and to the left from O, through A3 straight down from O, through A4 to the right and down from O, all the way to A5, straight to the right from O. For simplicity things like air resistance etc that reduce the mechanical energy of the system are ignored so unless disturbed it would keep oscillating between A1 and A5 for now.

What is the acceleration of the weight at points A1 .. A5? I'm especially interested in the direction of the acceleration but if one gets magnitude too, great.
2. 25 Oct '11 08:24
A quick search online (for the sake of curiosity given I'm not a mathematician) confirms I know absolutely nothing about circular momentum. I would have intuited the instantaneous (or average) acceleration at (or around) points A1 and A5 would be zero......but the 'web and the wording of your question suggests otherwise.....and that I don't understand acceleration as much as I thought I did.

Pass.
🙂
3. 25 Oct '11 19:122 edits
Originally posted by andrew93
A quick search online (for the sake of curiosity given I'm not a mathematician) confirms I know absolutely nothing about circular momentum. I would have intuited the instantaneous (or average) acceleration at (or around) points A1 and A5 would be zero......but the 'web and the wording of your question suggests otherwise.....and that I don't understand acceleration as much as I thought I did.

Pass.
🙂
On the contrary, at points A1 and A5, the pendulum's instantaneous velocity should be 0, and the pendulum is starting a freefall at -9.8m/s^2 until the string exerts force on it.

At any given point, the acceleration of the pendulum should be the sum of the two vectors of gravitational acceleration (g) and g * cos(theta) in the direction of the pin, where theta is the angle between the string and vertical.

I believe this results in an instantaneous acceleration of g * sin(theta) along the tangent to the arc.
4. 26 Oct '11 10:26
You've neglected the centripetal acceleration. This is v^2/r, which (by conservation of energy) is 2g.sin(theta).

So you've got the sum of g downwards, and 3g.sin(theta) towards the pin.

That gives you a magnitude of (3.sin^2(theta) + 1)*g, I think.
5. 26 Oct '11 14:08
As the square of sin(theta) oscillates between zero and 1, the above expression goes up and down between 3g and 4g?

6. 27 Oct '11 10:201 edit