Force of Gravitation and Centripetal Force

[Taniaz]

Homework Statement

(a) The Earth may be considered to be a uniform sphere of radius 6.37 × 103 km with its mass of 5.98 × 1024 kg concentrated at its centre. The Earth spins on its axis with a period of 24.0 hours. (i) A stone of mass 2.50 kg rests on the Earth’s surface at the Equator.

1. Calculate, using Newton’s law of gravitation, the gravitational force on the stone.

2. Determine the force required to maintain the stone in its circular path.

(ii) The stone is now hung from a Newton-meter.
Use your answers in (i) to determine the reading on the meter. Give your answer to three significant figures.

Homework Equations

F= GMm/r^2 and F= mw^2r

The Attempt at a Solution

I got the solution for 1. 24.6 N and 2. 0.0842 N but I don't understand why is (ii) or the weight the difference between these two values? Why do we have to subtract Fg-Fc, aren't they supposed to be the same thing?

 

[CWatters]

You have correctly calculated the gravitational force on the stone (24.6N) and the centripetal force required to maintain the stone in it's circular path (0.084N).

As for your question...

Why do we have to subtract Fg-Fc, aren't they supposed to be the same thing?

First thing to understand is that F_c is not a force that acts on the stone.

The calculation F_c = mv²/r gives you the net force (eg the sum of all other forces) required in order for the stone to move in a circle of the required radius and velocity.

There are two forces that act on the stone. One of the forces acting on the stone is Fg. What is the other force acting on the stone? Hint: Draw a free body diagram of the stone.

 

[Taniaz]

You have correctly calculated the gravitational force on the stone (24.6N) and the centripetal force required to maintain the stone in it's circular path (0.084N).

As for your question...
First thing to understand is that F_c is not a force that acts on the stone.

The calculation F_c = mv²/r gives you the net force (eg the sum of all other forces) required in order for the stone to move in a circle of the required radius and velocity.

There are two forces that act on the stone. One of the forces acting on the stone is Fg. What is the other force acting on the stone? Hint: Draw a free body diagram of the stone.

The other force is the weight of the stone?

 

[cnh1995]

The other force is the weight of the stone?

Look up 'normal reaction'. It often shows up in the FBDs of this type of situations where you have one body placed on another body (or surface).

 

[Taniaz]

Oh yes sorry! So how does that help us?

 

[cnh1995]

Oh yes sorry! So how does that help us?

It is one of the forces acting on the stone. Did you draw the FBD?

 

[Taniaz]

It is one of the forces acting on the stone. Did you draw the FBD?

Ok so the normal reaction is acting perpendicular to the surface of the Earth where the stone has been placed and Fg is acting downwards and Fc (net force of the two) is inwards?

 

[cnh1995]

Ok so the normal reaction is acting perpendicular to the surface of the Earth where the stone has been placed and Fg is acting downwards and Fc (net force of the two) is inwards?

Yes.

 

[Taniaz]

But I still don't understand why weight = Fg - Fc?

 

[cnh1995]

But I still don't understand why weight = Fg - Fc?

It's what is called the 'apparent weight' of the body. It is less than Fg (mg).

 

[Taniaz]

I don't get it, why is it minus Fc? What did normal reaction have to do with this? How did you know for the last part of the question we had to calculate the apparent weight of the stone?

 

[cnh1995]

I don't get it, why is it minus Fc?

The true weight of the stone, if the Earth suddenly stopped rotating, would be simply Fg(=mg) and the normal reaction would be equal to the true weight. But on the rotating earth, in order to maintain the circular motion of the stone, the normal reaction is less than the true weight of the stone such that the difference between them, or the net force on the stone is the necessary centripetal force.

How did you know for the last part of the question we had to calculate the apparent weight of the stone?

I am not sure what the Newton-meter should read. If the stone is suspended, it is no longer in contact with the surface, so no normal reaction. But there will be tension in the string to which the stone is tied , which will act same as the normal force. So the Newton-meter should show the apparent weight.

 

[Taniaz]

The true weight of the stone, if the Earth suddenly stopped rotating, would be simply Fg(=mg) and the normal reaction would be equal to the true weight. But on the rotating earth, in order to maintain the circular motion of the stone, the normal reaction is less than the true weight of the stone such that the difference between them, or the net force on the stone is the necessary centripetal force.

I am not sure what the Newton-meter should read. If the stone is suspended, it is no longer in contact with the surface, so no normal reaction. But there will be tension in the string to which the stone is tied , which will act same as the normal force. So the Newton-meter should show the apparent weight.

Ok that makes sense thanks.

One last question, Fg and the normal reaction don't really act along the same line unless they're either at the top or the bottom of the circle. When they're on the sides, they make a 90 degrees angle with one another, then you can't do Fg-Fc can you? Because they're perpendicular to each other?

 

[CWatters]

I don't get it, why is it minus Fc? What did normal reaction have to do with this? How did you know for the last part of the question we had to calculate the apparent weight of the stone?

Ok step by step..

The first thing to note is that the stone is accelerating. Therefore the net force acting on the stone is not zero.

So we can draw our FBD..

The two forces acting on the stone are F_g and the normal force F_N so we can write that as...

F_g + F_N <> 0

In order to move in a circle the net force must provide the centripetal force F_c (=mv²/r) so...

F_g + F_N = F_c

Rearrange to give the normal force...

F_N = F_c - F_g

The weight call it F_W (as measured by the Newton-meter) is the same magnitude as the Normal force but has the opposite sign so..

F_W = - (F_c - F_g) = F_g - F_c

 

[CWatters]

One last question, Fg and the normal reaction don't really act along the same line unless they're either at the top or the bottom of the circle. When they're on the sides, they make a 90 degrees angle with one another, then you can't do Fg-Fc can you? Because they're perpendicular to each other?

I can imagine a situation where F_g and F_N aren't parallel. For example a stone near a mountain would experience gravitational force that wasn't perfectly vertical. Not to scale!..

In that case you might need to think about other forces such as Ff the force due to friction.

Your starting point would be as before, that the net force = F_c..

F_N + F_g + F_f = F_c

(vector addition obviously)

You then rearrange it in the same way as before to give F_N and hence F_W.

F_N = F_c - (F_g + F_f)

F_W = (F_g + F_f) - F_c

Friction F_f can be calculated from the horizontal component of F_g