# How is centripetal force involved here? (charged mass sliding down a conducting hemisphere)

• tellmesomething
tellmesomething
Homework Statement
A block of charge q and mass m is kept on a smooth non conducting hemisphere . When it was released from rest it was found to lose contact either the hemisphere when the angle with the vertical was 37°. find mg/qE

The electric field is perpendicular to the gravity
Relevant Equations
None
I have some questions...
Firstly I drew the FBD of the said block at an angle theta from the vertical

Which force causes the block to lose contact.ie the normal to become zero...
Is it because the forces in the horizontal direction surpass some limit value of the net force in the radial direction...? I find this very confusing

I understand that the net force in the vertical direction would not be equal to zero when the block is in contact with the hemisphere....if it were 0 then the block would move in a straight line tangent to any point on the hemisphere....This net force is centripetal Force

Mathematically when the normal force equals zero the sum of the vertical forces would equal to the centripetal force..

But physically I cannot visualise it...its all very confusing... Is there a way I can back tract from the last equation, sum of vertical forces equals the net force m×a and find out a to be v²/r......?

Last edited:
Forces are the means by which one object communicates to another that it's there. There are two objects exerting one force each on the block.
1. The Earth, which exerts the force of gravity.
2. The hemisphere, which exerts the normal force.
If one of the forces goes to zero it's as if the object that generates it doesn't exist because it can longer communicate that it's there. "Losing contact" means that, as far as the block is concerned, only gravity exists and the block isn't there. It's as simple as that.

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kuruman said:
Forces are the means by which one object communicates to another that it's there. There are two objects exerting one force each on the block.
1. The Earth, which exerts the force of gravity.
2. The hemisphere, which exerts the normal force.
If one of the forces goes to zero it's as if the object that generates it doesn't exist because it can longer communicate that it's there. "Losing contact" means that, as far as the block is concerned, only gravity exists and the block isn't there. It's as simple as that.
But why is it losing contact is it because a force in the perpendicular direction increases? causing the resultant vector to move out of the trajectory of the circle say a non uniform circular motion?

tellmesomething said:
I

But why is it losing contact is it because a force in the perpendicular direction increases? causing the resultant vector to move out of the trajectory of the circle say a non uniform circular motion?
And if this is the case why did the block follow the trajectory of a circle till 37° why did it not move away ..? As the velocity was never constant which means there was a tangential acceleration at all times..

After reading the roblem more carefully, it seems that the block is charged. It makes sense because the angle would not be 37° if gravity and the normal force were the only forces. Please provide the exact statement of the problem as was given to you.
tellmesomething said:
But why is it losing contact is it because a force in the perpendicular direction increases? causing the resultant vector to move out of the trajectory of the circle say a non uniform circular motion?
You should be able to see that if you draw a FBD diagram, find an expression for the normal force and see what must be true for it to be zero.

kuruman said:
After reading the roblem more carefully, it seems that the block is charged. It makes sense because the angle would not be 37° if gravity and the normal force were the only forces. Please provide the exact statement of the problem as was given to you.

You should be able to see that if you draw a FBD diagram, find an expression for the normal force and see what must be true for it to be zero.

For the normal force I get this expression mgcostheta - E sin theta - mv²/r = N

But I dont understand how this tells me anything...I only get that the sum of the vertical forces equal the centripetal force when normal becomes 0

tellmesomething said:
For the normal force I get this expression mgcostheta - E sin theta - mv²/r = N

But I dont understand how this tells me anything...I only get that the sum of the vertical forces equal the centripetal force when normal becomes 0
First of all, the sum of all the vertical forces can only be a force along the vertical, in the direction of (or opposite to) the acceleration of gravity. The centripetal force by definition points towards the center of the hemisphere. You can only set the sum of the vertical forces equal to the centripetal force when the block sits at the very top of the hemisphere.

Secondly, you will need to find ##mv^2## assuming that the block starts from rest at the top of the hemisphere. That's an assumption that I have to make because you still have not provided a statement of the problem. How do you think you can find ##mv^2~##?

kuruman said:
First of all, the sum of all the vertical forces can only be a force along the vertical, in the direction of (or opposite to) the acceleration of gravity. The centripetal force by definition points towards the center of the hemisphere. You can only set the sum of the vertical forces equal to the centripetal force when the block sits at the very top of the hemisphere.

Secondly, you will need to find ##mv^2## assuming that the block starts from rest at the top of the hemisphere. That's an assumption that I have to make because you still have not provided a statement of the problem. How do you think you can find ##mv^2~##?
Well by vertical force I meant the vertical components of mg and electrostatic force.

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kuruman said:
First of all, the sum of all the vertical forces can only be a force along the vertical, in the direction of (or opposite to) the acceleration of gravity. The centripetal force by definition points towards the center of the hemisphere. You can only set the sum of the vertical forces equal to the centripetal force when the block sits at the very top of the hemisphere.

Secondly, you will need to find ##mv^2## assuming that the block starts from rest at the top of the hemisphere. That's an assumption that I have to make because you still have not provided a statement of the problem. How do you think you can find ##mv^2~##?
Also the homework statement specifically mentioned when the block is released from rest...

tellmesomething said:
Well by vertical force I meant the vertical components of mg and electrostatic force.
The electrostatic force appears not to have a vertical component from the looks of your drawing. You probably meant the sum of the weight and the electrostatic force. The centripetal force then would be the vector sum of (a) the weight, (b) the electrostatic force and (c) the normal force. Can you find an expression for that?

I dont understand what you're asking...? How can sum of weight and electrostatic force and normal be centripetal force.....
It is the sum of the vertical component of weight,electrostatic force and normal force which is equal to the centripetal force... Vertical component as in E sin theta and mg cos theta..
The expression being mg cos theta - E sin theta - N = mv²/r

kuruman said:
The electrostatic force appears not to have a vertical component from the looks of your drawing. You probably meant the sum of the weight and the electrostatic force. The centripetal force then would be the vector sum of (a) the weight, (b) the electrostatic force and (c) the normal force. Can you find an expression for that?
Edit : I just noticed that you said vector sum..sorry producing another expression...

kuruman said:
The electrostatic force appears not to have a vertical component from the looks of your drawing. You probably meant the sum of the weight and the electrostatic force. The centripetal force then would be the vector sum of (a) the weight, (b) the electrostatic force and (c) the normal force. Can you find an expression for that?
Wait doss that mean the net force is centripetal force.....? Tht doesnt make quite sense...that would

That wouldn't make quite sense... If I add these vectorially the resultant would not be in the direction of centripetal force

tellmesomething said:
Edit : I just noticed that you said vector sum..sorry producing another expression...
No, that's OK. It's the component in the centripetal direction that counts in this case.

tellmesomething said:
The expression being mg cos theta - E sin theta - N = mv²/r
This is incorrect. Check the electrical force.
tellmesomething said:
That wouldn't make quite sense... If I add these vectorially the resultant would not be in the direction of centripetal force
No. It would be in the direction of the net force. There is also a tangential component. You need to concentrate on the centripetal component because that's where the normal force appears.

So how are you going to find ##mv^2~##?

kuruman said:
No, that's OK. It's the component in the centripetal direction that counts in this case.

This is incorrect. Check the electrical force.

No. It would be in the direction of the net force. There is also a tangential component. You need to concentrate on the centripetal component because that's where the normal force appears.

So how are you going to find ##mv^2~##?
Oh sorry It should be E q sin theta

For mv² I would have to find the velocity at the point when it loses contact I can technically find the displacement but im contemplating whether the net acceleration is constant for using the SUVAT equations..

tellmesomething said:
Oh sorry It should be E q sin theta

For mv² I would have to find the velocity at the point when it loses contact I can technically find the displacement but im contemplating whether the net acceleration is constant for using the SUVAT equations..
Alternatively I could use WET but SUVAT equations are more intuitive..

tellmesomething said:
For mv² I would have to find the velocity at the point when it loses contact I can technically find the displacement but im contemplating whether the net acceleration is constant for using the SUVAT equations..
The SUVAT equations are applicable only when the acceleration is constant. That is not the case here. Hint: I wrote ##mv^2## instead of ##v## for a reason. Does this suggest something to you? Remember that the hemisphere is smooth.

kuruman said:
The SUVAT equations are applicable only when the acceleration is constant. That is not the case here. Hint: I wrote ##mv^2## instead of ##v## for a reason. Does this suggest something to you? Remember that the hemisphere is smooth.
Well yea I could use the work energy theorem .. The acceleration is Not constant because of the normal force right?

tellmesomething said:
Well yea I could use the work energy theorem .. The acceleration is Not constant because of the normal force right?
Then do it. The acceleration is not constant because it depends on ##\theta##.

kuruman said:
Then do it. The acceleration is not constant because it depends on ##\theta##.
Yes and the normal force depends on ## \theta ##

kuruman said:
Then do it. The acceleration is not constant because it depends on ##\theta##.
QE R sin theta+ mg (R- R cos theta) = 1/2 mv²-0

tellmesomething said:
Yes and the normal force depends on ## \theta ##
Could i not find a relation for theta and time t.... Since its tangential acceleration... Its probably equal to R Alpha so ## R \frac {d² \theta } {dt²} ## and then equal it to the expression I would get from the horizontal components..then integrate..

tellmesomething said:
Could i not find a relation for theta and time t.... Since its tangential acceleration... Its probably equal to R Alpha so ## R \frac {d² \theta } {dt²} ## and then equal it to the expression I would get from the horizontal components..then integrate..
You don't need to go there. You have
tellmesomething said:
The expression being mg cos theta - E sin theta - N = mv²/r
and you have
tellmesomething said:
QE R sin theta+ mg (R- R cos theta) = 1/2 mv²-0
What must be true for N to be equal to zero when θ = 37° ?

kuruman said:
You don't need to go there. You have

and you have

What must be true for N to be equal to zero when θ = 37° ?
Mg cos theta - E sin theta = mv²/r

tellmesomething said:
Mg cos theta - E sin theta = mv²/r
Probably need to divide these two equations
kuruman said:
You don't need to go there. You have
but what I meant is could this be an alternate solution...???

tellmesomething said:
Probably need to divide these two equations

but what I meant is could this be an alternate solution...???
I would take ##mv^2## from the work-energy equation and put it in the centripetal equation.

There is no alternate solution. Writing the work-energy energy equation is the way to approach this problem. Don't forget that the acceleration is a function of θ and θ is a function of t. You can say "I can integrate", but if want to find the velocity, you need to do an integral of the form $$v(t)=\int a(\theta(t))dt.$$Good luck with that.

tellmesomething
kuruman said:
I would take ##mv^2## from the work-energy equation and put it in the centripetal equation.

There is no alternate solution. Writing the work-energy energy equation is the way to approach this problem. Don't forget that the acceleration is a function of θ and θ is a function of t. You can say "I can integrate", but if want to find the velocity, you need to do an integral of the form $$v(t)=\int a(\theta(t))dt.$$Good luck with that.
OK. Thankyou for your replies, you help always.

kuruman
tellmesomething said:
Could i not find a relation for theta and time t.... Since its tangential acceleration... Its probably equal to R Alpha so ## R \frac {d² \theta } {dt²} ## and then equal it to the expression I would get from the horizontal components..then integrate..
It often happens that there is an analytic way to find velocity as a function of position (e.g. by energy conservation) but not as a function of time. In such cases, the substitution ##\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v## can be useful.
kuruman said:
The centripetal force then would be the vector sum of (a) the weight, (b) the electrostatic force and (c) the normal force.
To be exact, the component of that vector sum normal to the velocity is the centripetal force.

tellmesomething
haruspex said:
To be exact, the component of that vector sum normal to the velocity is the centripetal force.
Yes, of course. I had in mind the sum of the forces in the radial direction which is where the normal force appears.

kuruman said:
Yes, of course. I had in mind the sum of the forces in the radial direction which is where the normal force appears.
It can be surprising, though. On the thread about a planet in an elliptical orbit it struck me that the centripetal force is only towards the star at the apogee and perigee. At other points the gravitational attraction has a tangential component.

haruspex said:
It can be surprising, though. On the thread about a planet in an elliptical orbit it struck me that the centripetal force is only towards the star at the apogee and perigee. At other points the gravitational attraction has a tangential component.
Sure. This is similar to 2D projectile motion near the surface of the Earth. There is always a centripetal component that becomes equal to the weight at the top of the trajectory.

haruspex said:
It often happens that there is an analytic way to find velocity as a function of position (e.g. by energy conservation) but not as a function of time. In such cases, the substitution ##\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v## can be useful.

To be exact, the component of that vector sum normal to the velocity is the centripetal force.
I see... Thankyou :)

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