[Yorrap]

Please confirm the following points and fix any mistakes!

1- An object moving in a circular path must have a centripetal force acting on it.
2- Centripetal force is the sum of all forces acting on an object moving in a circle.

3- Example A:
Car is driven over a hump back bridge of radius, r with constant speed, v.
In this case Centripetal Force = Weight (mg) + Normal Reaction (-R)

4- If the same car in example A is driven with an increasing speed, it will lose the contact with the road for a while and travel in a tangential direction to the circular path because the the centripetal force needed to maintain the car in this circular path is now greater than the combined forces ( mg + (-R) ) acting on the car. (Is this correct?)

Just like this;




5- Example B
A satellite orbits the earth with a constant speed, so the only force acting on the satellite is the gravitational pull of the Earth and centripetal force needed to maintain this circular motion is provided by the gravity, which is equal to the weight of the satellite.

So, Centripetal Force (mv²/r) = Weight (mg)

- This means, the satellite must have a (minimum) constant speed in order to maintain the circular motion. If it decelerates and speed is "below a certain" value, the satellite will fall towards the Earth because the weight is greater than the centripetal force.

- If it accelerates and speed exceeds a "certain" value, it will travel in a straight line because the centripetal force needed can no longer be provided by the gravity.

(Is this correct?)

[Ari Ben Canaan]

Example B - I'm not comfortable describing the centripetal force as equal to the weight of the object.

You seem to be implying that the object has constant weight at all points at a distance from the earth when in fact the gravitational field strength and hence the 'weight' of the object decrease as per an inverse square law.

Hence, I would describe it like this :

Centripetal force on satellite at distance R from earth :

mv^2/r = GMm/r^2 I've used Newton's Law of Gravitation

v^2=GM/r

Hence, it becomes clear that decreasing the velocity of a satellite in geostationary object will cause the gravitational force of attraction to pull the satellite downwards since the left hand side of the equation is lower than the right hand side.

[Stonebridge]

4. The path of the car if it leaves the road will be a parabola. It's the same trajectory that a projectile follows if projected at an angle equal to that which the car was travelling just as it left the road. This is the result of an acceleration of g vertically downwards and a constant component of velolocity in the horizontal direction. (Ignoring air resistance.)

[Yorrap]

Thanks for the quick replies!

(Original post by Ari Ben Canaan)
Example B - I'm not comfortable describing the centripetal force as equal to the weight of the object.

You seem to be implying that the object has constant weight at all points at a distance from the earth when in fact the gravitational field strength and hence the 'weight' of the object decrease as per an inverse square law.

Hence, I would describe it like this :

Centripetal force on satellite at distance R from earth :

mv^2/r = GMm/r^2 I've used Newton's Law of Gravitation

v^2=GM/r

Hence, it becomes clear that decreasing the velocity of a satellite in geostationary object will cause the gravitational force of attraction to pull the satellite downwards since the left hand side of the equation is lower than the right hand side.

Sorry! Seems like my writing "g" for the new value and calling it "weight" is a huge mistake here because i am not actually referring to g = 9.81Nkg^-1 and weight = m x 9.81Nkg^-1

Would it be wrong to say Centripetal Force = mg_new ,if let's say the radius of this satellite's circular path is 35 000 km. The radius is constant and the gravitational field strength acting on the satellite has to be constant aswell, as long as the radius stays the same.

so new gravitational field strength, g_new = GM_E/r²

g_new = 6.67x10^-11 Nm²kg^-2 x 5.9722 × 10²⁴ kg / (35 000 x 1000)m = 0.33 Nkg^-1

So the Centripetal Force = Gravitational Force acting on the satellite, is constant as long as the radius remains constant.

(Original post by Stonebridge)
4. The path of the car if it leaves the road will be a parabola. It's the same trajectory that a projectile follows if projected at an angle equal to that which the car was travelling just as it left the road. This is the result of an acceleration of g vertically downwards and a constant component of velolocity in the horizontal direction. (Ignoring air resistance.)

Then in the case of the satellite, it will be a straight line? (red path)

[Stonebridge]

(Original post by Yorrap)
Thanks for the quick replies!



Sorry! Seems like my writing "g" for the new value and calling it "weight" is a huge mistake here because i am not actually referring to g = 9.81Nkg^-1 and weight = m x 9.81Nkg^-1

Would it be wrong to say Centripetal Force = mg_new ,if let's say the radius of this satellite's circular path is 35 000 km. The radius is constant and the gravitational field strength acting on the satellite has to be constant aswell, as long as the radius stays the same.

so new gravitational field strength, g_new = GM_E/r²

g_new = 6.67x10^-11 Nm²kg^-2 x 5.9722 × 10²⁴ kg / (35 000 x 1000)m = 0.33 Nkg^-1

So the Centripetal Force = Gravitational Force acting on the satellite, is constant as long as the radius remains constant.




Then in the case of the satellite, it will be a straight line? (red path)

Possibly but not necessarily. In general if it were to speed up its path would become straighter. For it to become straight the rocket booster used would have to apply a force that exactly cancelled the sideways (originally centripetal) force of attraction of the Earth on the satellite.
So it's more complicated than it first seems.
The general answer would be straighter (=larger radius of curvature) rather than straight if the satellite were to somehow accelerate.

[trustmeoff]

Hey guys I'm planning to do Physics A'level on my own and wanted to ask which book has the most resources and is useful when you're studying without a tutor. Edexcel (Longman), Edexcel (Hodder education) or salters horners Advanced physics. Thanks

[Ari Ben Canaan]

(Original post by trustmeoff)
Hey guys I'm planning to do Physics A'level on my own and wanted to ask which book has the most resources and is useful when you're studying without a tutor. Edexcel (Longman), Edexcel (Hodder education) or salters horners Advanced physics. Thanks

Are you doing this to answer an exam or out if interest ?

This was posted from The Student Room's Android App on my GT-I9100

[Yorrap]

(Original post by Stonebridge)
Possibly but not necessarily. In general if it were to speed up its path would become straighter. For it to become straight the rocket booster used would have to apply a force that exactly cancelled the sideways (originally centripetal) force of attraction of the Earth on the satellite.
So it's more complicated than it first seems.
The general answer would be straighter (=larger radius of curvature) rather than straight if the satellite were to somehow accelerate.

Here is an exam question and a sentence from examiner report;

If centripetal force is supplied by weight, what is meant by "so long as the required force was greater than weight..." ? How can centripetal force be greater than the weight, if weight is supplying it?

Is this another way of saying "as long as it is swung with a force equal to weight or greater than weight, it will not fall?

And what happens to Reaction from bucket here? Is it zero? or is it a non zero value and they are only looking for the "idea" that minimum mv²/r should be equal to = mg? (where it is mv²/r = mg + R here)




Mark Scheme

Spoiler:

Show

(Original post by trustmeoff)
Hey guys I'm planning to do Physics A'level on my own and wanted to ask which book has the most resources and is useful when you're studying without a tutor. Edexcel (Longman), Edexcel (Hodder education) or salters horners Advanced physics. Thanks

I haven't used any of these. I use these two; (Edexcel Students Book)

(and pretty useless in my opinion, lol).

(Original post by Ari Ben Canaan)
Are you doing this to answer an exam or out if interest ?

This was posted from The Student Room's Android App on my GT-I9100

comments on this?
http://www.thestudentroom.co.uk/show...5#post37834225

[Stonebridge]

(Original post by Yorrap)
Here is an exam question and a sentence from examiner report;

If centripetal force is supplied by weight, what is meant by "so long as the required force was greater than weight..." ? How can centripetal force be greater than the weight, if weight is supplying it?

Is this another way of saying "as long as it is swung with a force equal to weight or greater than weight, it will not fall?

And what happens to Reaction from bucket here? Is it zero? or is it a non zero value and they are only looking for the "idea" that minimum mv²/r should be equal to = mg? (where it is mv²/r = mg + R here)

[


comments on this?
http://www.thestudentroom.co.uk/show...5#post37834225

At the top of the bucket's motion there are two forces that can actually act on the water
its weight (mg downwards)
the reaction of the bottom of the bucket (R downwards)

What the examiner is saying is that if the water is to perform circular motion it needs a centripetal force, at that point, a DOWNWARDS force and equal to mv²/r

Where is it going to come from?
mg and R

So mg and R together must be equal to mv²/r

If the weight of the object is sufficient then there doesn't need to be any R.
mg provides the whole of the centripetal force and R=0
here mg = mv²/r

If you swing the bucket faster then mv²/r gets larger and mg isn't sufficient as it doesn't change.
In this case R then has to provide the rest of the centripetal force
Then mg + R =mv²/r


So long as the required centripetal force is equal to or greater than the weight of the water, it will stay in the bucket. This is the two cases described above.
If mg is greater than the required centripetal force, the water will start to fall out of the bucket.

[Yorrap]

(Original post by Stonebridge)
At the top of the bucket's motion there are two forces that can actually act on the water
its weight (mg downwards)
the reaction of the bottom of the bucket (R downwards)

What the examiner is saying is that if the water is to perform circular motion it needs a centripetal force, at that point, a DOWNWARDS force and equal to mv²/r

Where is it going to come from?
mg and R

So mg and R together must be equal to mv²/r

If the weight of the object is sufficient then there doesn't need to be any R.
mg provides the whole of the centripetal force and R=0
here mg = mv²/r

If you swing the bucket faster then mv²/r gets larger and mg isn't sufficient as it doesn't change.
In this case R then has to provide the rest of the centripetal force
Then mg + R =mv²/r


So long as the required centripetal force is equal to or greater than the weight of the water, it will stay in the bucket. This is the two cases described above.
If mg is greater than the required centripetal force, the water will start to fall out of the bucket.

Now this cleared my confusion!

I completely understood (and knew) the green part but the blue part contains new information.

I didn't know that R becomes zero, if weight, mg supplies the whole centripetal force and in cases if it can't provide the needed centripetal force that reaction force, R provides the needed force.

So this part of the sentence "as long as the speed of the bucket exceeds a certain value" should make me realise that Reaction, R must be zero so that weight can be equal to the centripetal force (at minimum velocity) to maintain the circular motion, right?


Is this diagram correct for the same bucket at the bottom?

Spoiler:

Show

This is the wrong diagram!

Question : If diagram is correct, does this mean that R can never become zero at the bottom?

[Stonebridge]

Yes I think you've got it now.
In your diagram with the bucket at the bottom, the reaction R is the only force that can provide the necessary upwards centripetal force. It acts upwards but the weight acts downwards.
So here the equation is R - mg = mv²/r
If R is positive (up), mg is negative (down)
The centripetal force is up so positive.
R can never be zero because R= mv²/r + mg


BTW
If you say mg is positive (down is positive as you did in your diagram) and R is negative (opposite direction)
the centripetal force is also negative as it's upwards.

The equation becomes
-R + mg = -mv²/r

This is, of course, exactly equivalent to the other one but with all the signs reversed.
Multiply both sides by -1 to get the other equation.

[Yorrap]

(Original post by Stonebridge)
Yes I think you've got it now.
In your diagram with the bucket at the bottom, the reaction R is the only force that can provide the necessary upwards centripetal force. It acts upwards but the weight acts downwards.
So here the equation is R - mg = mv²/r
If R is positive (up), mg is negative (down)
The centripetal force is up so positive.
R can never be zero because R= mv²/r + mg


BTW
If you say mg is positive (down is positive as you did in your diagram) and R is negative (opposite direction)
the centripetal force is also negative as it's upwards.

The equation becomes
-R + mg = -mv²/r

This is, of course, exactly equivalent to the other one but with all the signs reversed.
Multiply both sides by -1 to get the other equation.

Yes, downwards positive and upwards negative.
I edited my earlier post and added the correct diagram.


I think that's all they can ask about circular motion in a vertical circle. Top and bottom.