Potential energy of a ball on top of a vertical rod

[Vriska]

Homework Statement

The problem is to take a small ball of mass m on a uniform rod of mass m which is hung on a hinge vertically downward, they're asking to find the velocity to be imparted for it undergo a complete rotation.

I have the kinetic energy bit worked out but I'm getting the wrong answer equating it to potential energy

Homework Equations

Cm = x1m1+x2m2/(m1+m2)

The Attempt at a Solution

[/B]
Okay I'll assume the ball system nearly comes to a stop at the when it stands vertically. the cm of the rod will be l/2. this means that the mass is concentrated at l/2. now cm of the total system would be half way between the balland the cm of the rod. So it's l/2 + l/4 which is 3/4l. Mass = 2m so potential energy is lmg3/2.

.The books says this energy is 3mgl

 

[ehild]

Homework Statement

The problem is to take a small ball of mass m on a uniform rod of mass m which is hung on a hinge vertically downward, they're asking to find the velocity to be imparted for it undergo a complete rotation.

I have the kinetic energy bit worked out but I'm getting the wrong answer equating it to potential energy

Homework Equations

Cm = x1m1+x2m2/(m1+m2)

The Attempt at a Solution

[/B]
Okay I'll assume the ball system nearly comes to a stop at the when it stands vertically. the cm of the rod will be l/2. this means that the mass is concentrated at l/2. now cm of the total system would be half way between the balland the cm of the rod. So it's l/2 + l/4 which is 3/4l. Mass = 2m so potential energy is lmg3/2.

.The books says this energy is 3mgl

You are right.

 

[Vriska]

You are right.

Ah thank you for confirming my suspicions, this problems was a bugger

 

[CWatters]

Okay I'll assume the ball system nearly comes to a stop at the when it stands vertically. the cm of the rod will be l/2. this means that the mass is concentrated at l/2. now cm of the total system would be half way between the balland the cm of the rod. So it's l/2 + l/4 which is 3/4l. Mass = 2m so potential energy is lmg3/2.

The books says this energy is 3mgl

I think the book is right. The cm moves from -3l/4 to +3l/4 a height gain of 3l/2. The mass is 2m so the gain in PE is

2m * g * 3l/2 = 3mgl

 

[Vriska]

I think the book is right. The cm moves from -3l/4 to +3l/4 a height gain of 3l/2. The mass is 2m so the gain in PE is

2m * g * 3l/2 = 3mgl

?! woah, thank you so much for correcting me

 

[PeroK]

You can also simply take the ball and the rod separately:

The ball moves upward by $2l$, so its gain in PE is $2lmg$.

The COM of the rod is $l/2$ below the hinge, so its gain in PE is $lmg$.

The total gain in PE, therefore, is $3lmg$.