Solving for Maximum Speed of Ball in a Pendulum Swing

[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?

 

[dextercioby]

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

Daniel.

 

[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.

 

[dextercioby]

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

Daniel.

 

[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 isn't angular velocity?

 

[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.

 

[UrbanXrisis]

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

 

[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 ) 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.