Vertical Circular Motion Problem

[Mickey Tee]

Homework Statement

Homework Equations

Centripetal Acceleration = v²/r
F_net = ΣF

The Attempt at a Solution

The block exerts a force W equal to its weight on the box at the top most position.
This upward W minus W downwards and an unknown F downwards equals a net F_c downwards. So F equals F_c

At the bottom position there is a W downwards and the total force on the box is this W plus something else
The net force is F_c upwards since the speed is constant

Now I get completely confused.

 

[ehild]

Homework Statement

View attachment 76215

Homework Equations

Centripetal Acceleration = v²/r
F_net = ΣF

The Attempt at a Solution

The block exerts a force W equal to its weight on the box at the top most position.This upward W minus W downwards and an unknown F downwards equals a net F_c downwards. So F equals F_c

What is the direction of the force the stone applies at the box? What is the force the box exerts on the stone? So what is your unknown force?

At the bottom position there is a W downwards and the total force on the box is this W plus something else
The net force is F_c upwards since the speed is constant

Now I get completely confused.

That "something else " is the normal force from the block, the block exerts on the stone.

 

[Mickey Tee]

What if I think in terms of pseudo force? The pseudo force at the top position is the centripetal force upwards, so the force exerted by the stone on the box is
F_c - mg = mg
=> F_c = 2mg

At the bottom position, pseudo force is downwards, along with mg. So the force by the stone on the box is 3mg. But the answer is 7mg :(

 

[ehild]

What if I think in terms of pseudo force? The pseudo force at the top position is the centripetal force upwards, so the force exerted by the stone on the box is
F_c - mg = mg
=> F_c = 2mg

At the bottom position, pseudo force is downwards, along with mg. So the force by the stone on the box is 3mg. But the answer is 7mg :(

You make yourself confused with the pseudo force. It is not the centripetal force. Use inertial frame of reference. The stone moves together with the box along a circle. It needs the appropriate centripetal force. The centripetal force is the resultant of two forces, what are they at the top of the circle?

 

[Mickey Tee]

I think at the top its (mg - N). But isn't there also an mg downwards? And isn't the normal force self adjusting?

Um, can you tell me how the normal force varies as the box goes around the circle?

 

[haruspex]

isn't the normal force self adjusting?

Yes, it adjusts as necessary to keep the stone moving in the circle - as long as that does not involve its going negative.

The centripetal force is the resultant of two forces, what are they at the top of the circle?

I think at the top its (mg - N)

You think the resultant is that?

But isn't there also an mg downwards?

I think you may be getting confused by the happenstance that the normal force equals mg here. Set that aside for the moment and just label it N. Which way does N act at the top? Which way does the weight mg act at the top?

 

[ehild]

I think at the top its (mg - N). But isn't there also an mg downwards? And isn't the normal force self adjusting?

The picture shows the box touching the stone on the upper side. The box can only push the stone, in what direction? Up or down?

 

[Jazz]

You're not told that there are external forces acting on the circle. In addition to what has been already said, it can be helpful to think about it in terms of a conserved quantity.

I don't see a reason to assume that the velocity is constant as the problem says.

 

[ehild]

The normal force would be 3mg in case of constant speed. If case of conservation of energy, the normal force would be 7mg.