Monkey Climbing Rope: Finding the Torque and Angular Momentum

[UrbanXrisis]

The monkey climbs the rope trying to get the banana seen http://home.earthlink.net/~urban-xrisis/clip002.jpg

I need to find the net torque and the total angular momentum about the pully axis.

I am absolutly stumped but I'll give it a shot. So the monkey is moving upwards that means it has to sustain and move its own body. That means it has to apply a force of Mg to hold onto the rope and Mg to move up the rope. The total force is 2mg. Total torque would be the force of the monkey (2Mgr) plus the banana (Mg) which equals 3Mgr right?

For the angular momentum, that is the integral of torque. Does that mean angular momentum is \frac{3Mgr^2}{2}

any help?

 

[MathStudent]

from the description in the picture, it seems as if the monkey is just clinging, and not actually moving with respect to the rope. So this means that its just the monkeys weight that adds to the tension of the rope.

Torque is a vector quantity, so it matters which direction the torque tries to rotate the pulley. By convention, a torque that tries to rotate an object clockwise about a pivot is considered negative, (just like a clockwise angle rotation is negative), and converseley a torque that tries to rotate the object counter-clockwise is positive.

 

[Doc Al]

I am absolutly stumped but I'll give it a shot. So the monkey is moving upwards that means it has to sustain and move its own body. That means it has to apply a force of Mg to hold onto the rope and Mg to move up the rope. The total force is 2mg. Total torque would be the force of the monkey (2Mgr) plus the banana (Mg) which equals 3Mgr right?

This is incorrect.

If the monkey climbs at a constant speed, what must be the tension in the rope?

Note that you are asked to find the net torque about the pulley. Hint: Assume the tension is the same throughout the rope.

 

[UrbanXrisis]

there is Mgr and -Mgr so that means that the net torque is zero?

 

[Doc Al]

That is correct.

 

[UrbanXrisis]

then that means there is no angular momentum right?

 

[Doc Al]

No torque means no change in angular momentum. But, assuming things start from rest, then the net angular momentum will remain zero.

 

[UrbanXrisis]

so the monkey can climb up the rope without the pully rotating because of opposite torques makes the net torque zero and therefore not change in angular velocity. Then that means the monkey can reach the bananas right?

 

[UrbanXrisis]

http://www.geocities.com/activityworkshop/puzzlesgames/monkey/solution.html

"Now, what happens when the monkey tries to climb the rope? The monkey exerts an additional force on the rope, so it pulls the rope down with a force which is now greater than w."

Isn't this the opposite of what should happen? Does this mean there should be a net torque greater than zero?

 

[Doc Al]

so the monkey can climb up the rope without the pully rotating because of opposite torques makes the net torque zero and therefore not change in angular velocity. Then that means the monkey can reach the bananas right?

No. The rotation of the pulley, which is assumed to be very light, will not affect the angular momentum of the system. But the monkey climbing the rope surely will. If the monkey raises himself at speed v, what must happen to the bananas so that the net angular momentum of the system remains unchanged?

 

[Doc Al]

http://www.geocities.com/activityworkshop/puzzlesgames/monkey/solution.html

"Now, what happens when the monkey tries to climb the rope? The monkey exerts an additional force on the rope, so it pulls the rope down with a force which is now greater than w."

Isn't this the opposite of what should happen? Does this mean there should be a net torque greater than zero?

No. Any additional force that the monkey exerts is transmitted, via the light rope, to the bananas. As I said before, the tension in the rope is the same throughout.