Harmonic motion

1. Apr 12, 2005

UrbanXrisis

A 5 kg ball hangs from a 10 m strong. The ball is swung horizontally outward 90 degrees from its equilibrium position. Assuming the system behaves as a simple pendulum, find the maximum speed of the ball during its swing.

what would I have to do to figure this problem out?

$$\omega = \sqrt{\frac{g}{l}}$$
$$\omega = \sqrt{\frac{9.8}{10}}$$
$$\omega=0.99rad/s$$

$$\omega r =v$$
$$0.99rad/s* 10m =v$$
$$v=9.9m/s$$

I'm not getting the answer of 14, what am I doing wrong?

2. Apr 12, 2005

dextercioby

HINT:Use the law of conservation of total mechanical energy.

Daniel.

3. Apr 12, 2005

Physicsisfun2005

yes conservation of energy is always better than mechanics when it comes to fussy math equations. Think of the change in gravitation potential energy.

4. Apr 12, 2005

dextercioby

And BTW,$v=\omega R$ could work in this case if u knew the maximum angular velocity...

Daniel.

5. Apr 12, 2005

UrbanXrisis

okay, i used $$gh=.5v^2$$ and got the answer I was looking for

As for $$v=\omega R$$, isn't that what $$\omega=\sqrt{\frac{g}{l}}$$ is? what is omega in that previous equation if it isnt angular velocity?

6. Apr 12, 2005

dextercioby

Nope,angular velocity is a very complicated function (something involving elliptic functions "cn" and "dn"),because the linear approximation $\sin \vartheta\simeq \vartheta$ would not be correct...

Daniel.

7. Apr 12, 2005

UrbanXrisis

I read in the book that omega in $$\omega=\sqrt{\frac{g}{l}}$$ is angular frequency. How is that different from angular velocity?

8. Apr 12, 2005

dextercioby

Angular velocity is

$$\omega (t)=:\frac{d\vartheta (t)}{dt}$$

and angular frequency is

$$\omega =:\frac{2\pi}{T}$$

These 2 #-s (denoted the same :yuck:) are equal only for a uniform circular motion .The bob from a mathematical pendulum (not even in the linear approximation) doesn't have a uniform circular motion,ergo the two "animals" are different.

Daniel.

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