Deriving C=mv2/r and a=v2/r: Mass, Gravity & Acceleration

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In summary, the conversation discusses the derivation of the formula v2/r and the reasoning behind why the mass cancels out in certain equations. The conversation also explores the relationship between centripetal acceleration, tangential velocity, and angular velocity in a rotating body. The formula v2/r is derived using the concept of distance around a circle and tangential velocity, and the mass cancels out because the amount of centripetal force acting on a rotating body is independent of its mass. The conversation ends with a question about the exponent of 0.5 in the formula.
  • #1
fcb
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I don't know if this is the right equation but in a question where it sakes the acceleration is 60%of the cars weight. Why does the mass cancel each other out so in the end you don't need to know the mass?

Sorry I am typin this on my phone so I am sorry for the spelling errors


Thanks

Edit: some won understand what I mean. When they say the acceration is a certain percentage of the weight. Why do you multiply it but gravity. Essestially using the formula v2/r

Can someone derive it for me. How does v2/r come about? How is there mass in the first place and why do you multiply it with the gravity constant of earth?
 
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  • #2
I'm not entirely sure what you're asking..
From what I can gather, you want to know how to derive a=v2/r and why weight is equal to mg, correct?
 
  • #3
fcb said:
I don't know if this is the right equation but in a question where it sakes the acceleration is 60%of the cars weight. Why does the mass cancel each other out so in the end you don't need to know the mass?

Sorry I am typin this on my phone so I am sorry for the spelling errors


Thanks

Edit: some won understand what I mean. When they say the acceration is a certain percentage of the weight. Why do you multiply it but gravity. Essestially using the formula v2/r

Can someone derive it for me. How does v2/r come about? How is there mass in the first place and why do you multiply it with the gravity constant of earth?

Let's start with where they get V2/r from.
If you want to know the distance around some part of a circle given an angle theta you can find it by using S = theta*r, take the derivative of each side to get the tangential velocity going around the circle:

dS/dt = (dtheta/dt)*r = V​
and dtheta/dt is the angular velocity, w. So,

V = w*r​
Now you need units of acceleration on each side, this will be the centripetal acceleration that we find having units of m/s2, so square each side and divide by r yielding:

V2/r = w2r​
Now V2/r has units of m2/m*s2 = m/s2, which are the correct units. So the centripetal acceleration is given by two formulas, those being:

V2/r = w2r​

OK, now you have been given the centripetal acceleration as 60% of the cars weight so the centripetal force acting on the car which acts to pull it inward like a satellite orbiting the Earth must be: 0.6mg, now set that equal to the centripetal acceleration times the mass of the car and you should be all set:

0.6mg = m*V2/r​
This is where the mass cancels on each side. How much of a centripetal force there is acting on a rotating body is independent of the mass of that body and is dependent only on the tangential speed it is moving at and the distance from the origin of the circle it travels along. So,


(5.88*r)0.5 = V​
Tangential Velocity of Car
where 5.88 = 0.6*g = 0.6*9.8 = 5.88

Or,

0.6mg = m*w2r​
5.88 = w2r
(5.88/r)0.5 = w​
The angular velocity of the car

So if you need the tangential velocity of the car use:
(5.88*r)0.5 = V​
Or if you need the angular velocity (how many times per second does the car make a complete circle) then use:
(5.88/r)0.5 = w​

Hope this helps. :smile:
Craig
 
  • #4
clombard1973 said:
Let's start with where they get V2/r from.
If you want to know the distance around some part of a circle given an angle theta you can find it by using S = theta*r, take the derivative of each side to get the tangential velocity going around the circle:

dS/dt = (dtheta/dt)*r = V​
and dtheta/dt is the angular velocity, w. So,

V = w*r​
Now you need units of acceleration on each side, this will be the centripetal acceleration that we find having units of m/s2, so square each side and divide by r yielding:

V2/r = w2r​
Now V2/r has units of m2/m*s2 = m/s2, which are the correct units. So the centripetal acceleration is given by two formulas, those being:

V2/r = w2r​

OK, now you have been given the centripetal acceleration as 60% of the cars weight so the centripetal force acting on the car which acts to pull it inward like a satellite orbiting the Earth must be: 0.6mg, now set that equal to the centripetal acceleration times the mass of the car and you should be all set:

0.6mg = m*V2/r​
This is where the mass cancels on each side. How much of a centripetal force there is acting on a rotating body is independent of the mass of that body and is dependent only on the tangential speed it is moving at and the distance from the origin of the circle it travels along. So,


(5.88*r)0.5 = V​
Tangential Velocity of Car
where 5.88 = 0.6*g = 0.6*9.8 = 5.88

Or,

0.6mg = m*w2r​
5.88 = w2r
(5.88/r)0.5 = w​
The angular velocity of the car

So if you need the tangential velocity of the car use:
(5.88*r)0.5 = V​
Or if you need the angular velocity (how many times per second does the car make a complete circle) then use:
(5.88/r)0.5 = w​

Hope this helps. :smile:
Craig

That was so helpful. Honestly,Thanks so much
 
  • #5
Quick question. Why is it to the power of 0.5??
 

1. How do you derive the equation C=mv^2/r?

To derive the equation C=mv^2/r, we start with the basic definition of centripetal force, which is defined as the force that causes an object to move in a circular path. This force is equal to the mass of the object multiplied by its centripetal acceleration, which is equal to v^2/r. Therefore, the equation can be written as C=mv^2/r.

2. What is the relationship between mass and centripetal force?

The mass of an object is directly proportional to the centripetal force acting on it. This means that as the mass of an object increases, the centripetal force required to keep it moving in a circular path also increases.

3. How does gravity affect the equation C=mv^2/r?

Gravity does not directly affect the equation C=mv^2/r. However, gravity can play a role in providing the necessary centripetal force to keep an object moving in a circular path. For example, the force of gravity between the Earth and a satellite orbiting it provides the centripetal force needed to keep the satellite in its orbit.

4. What is the relationship between acceleration and centripetal force?

The acceleration of an object moving in a circular path is equal to v^2/r, where v is the velocity of the object and r is the radius of the circular path. This acceleration is also equal to the centripetal force divided by the mass of the object. Therefore, there is a direct relationship between acceleration and centripetal force, with an increase in acceleration resulting in an increase in centripetal force.

5. Can the equation a=v^2/r be used for any type of circular motion?

Yes, the equation a=v^2/r can be used for any type of circular motion, whether it be a simple circular path or a more complex spiral or elliptical path. As long as the object is moving in a circular path with a constant radius, this equation can be used to calculate the centripetal acceleration. However, if the radius of the circular path is changing, additional equations and concepts, such as calculus, may be needed to accurately calculate the acceleration.

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