Circular motion and centripetal force

In summary, a particle of mass m kg is displaced from the top of a smooth sphere of radius 2 m and slides down, leaving the sphere at point B and striking the table at point Q. The speed of the particle at point B can be found using the formula for centripetal force, F=mv^2/r, which assumes a constant speed and acceleration towards the center of the circle. However, the particle is accelerating under the influence of gravity, so this equation must be used instantaneously at point B. At this point, the normal component of gravity must equal mv^2/r in order to provide the necessary centripetal force. The particle then starts free-falling and its speed at Q can be found using conservation
  • #1
evansmiley
16
0

Homework Statement


A particle of mass m kg lies on the top of a smooth sphere of radius 2 m.
The sphere is fixed on a horizontal table at P.
The particle is slightly displaced and slides down the sphere.
The particle leaves the sphere at B and strikes the table at Q.
Find
(i) the speed of the particle at B
(ii) the speed of the particle on striking the table at Q.



Homework Equations


F = mv^2/r

The Attempt at a Solution


I had no way to go about this question (however part ii can be easily gotten from conversion from potential energy to kinetic), but it is from a section on circular motion, and in the marking scheme for this question, they write force equations for the mass as if it were undergoing circular motion ie they use the equation for centripetal force F = mv^2/r . The solution is at http://www.examinations.ie/archive/markingschemes/2010/LC020ALP000EV.pdf [Broken] for anyone interested, it's question 6 part a. I'm just questioning their use of the formula for circular motion - surely, the formula only holds if v is a constant, however, the mass is clearly accelerating so this is not the case, so how can they justify using this equation in their solution? Thanks.
 
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  • #2
The formula is being applied at point B, at that point, they are finding the value for v.

However, the unwritten assumption is that the speed is constant. Otherwise you'd have an additional acceleration term.


Remember you can travel in a circle at constant speed but still be accelerating because there is a change in velocity due to a change in direction.
 
  • #3
but it starts off at rest so to have any velocity it must have accelerated ie the magnitude of it's velocity increases?
 
  • #4
evansmiley said:
but it starts off at rest so to have any velocity it must have accelerated ie the magnitude of it's velocity increases?

It accelerated under the influence of gravity. Otherwise you'd have a radial component and a tangential component for acceleration. So it is more or less circular motion with constant speed.
 
  • #5
Although the speed is not constant while the particle is on the sphere, the equation for circular motion holds instantaneously. Note that as the particle moves, the tangential and radial components of the force are also changing.
 
  • #6
The point at which the particle leaves the sphere is when it starts free-fall.
Before this point the particle 'sits' on the sphere because the R>0, =>mgCosθ >mv^2/r.

Yes, the particle were undergoing circular motion.
 
  • #7
ok i kind of see what you mean, but would it would be right to say that this is not uniform circular motion because the angular velocity is not constant?
 
  • #8
evansmiley said:
ok i kind of see what you mean, but would it would be right to say that this is not uniform circular motion because the angular velocity is not constant?

It is not uniform circular motion, but it is circular motion while the particle is in the sphere (point B). While moving on he sphere, its radial component of acceleration is v^2/R. That means the resultant force on the particle must have mv^2/R radial component. What forces act on the particle?

ehild
 
  • #9
evansmiley said:
ok i kind of see what you mean, but would it would be right to say that this is not uniform circular motion because the angular velocity is not constant?

It is not uniform circular motion, but it is circular motion while the particle is on the sphere (point B). While moving on the sphere, the radial component of acceleration is v^2/R. That means the resultant force on the particle must have mv^2/R radial component. What forces act on the particle?

ehild
 
  • #10
ah ok. gravity and reaction?
 
  • #11
evansmiley said:
ah ok. gravity and reaction?

Yes. Note that the reaction force can not pull the particle, it pushes only.

Apply conservation of energy to get the speed of the particle at a certain position. Check if the normal component of gravity is sufficient to supply the centripetal force.

ehild
 
  • #12
evansmiley said:

Homework Statement


A particle of mass m kg lies on the top of a smooth sphere of radius 2 m.
The sphere is fixed on a horizontal table at P.
The particle is slightly displaced and slides down the sphere.
The particle leaves the sphere at B and strikes the table at Q.
Find
(i) the speed of the particle at B
(ii) the speed of the particle on striking the table at Q.



Homework Equations


F = mv^2/r

The Attempt at a Solution


I had no way to go about this question (however part ii can be easily gotten from conversion from potential energy to kinetic), but it is from a section on circular motion, and in the marking scheme for this question, they write force equations for the mass as if it were undergoing circular motion ie they use the equation for centripetal force F = mv^2/r . The solution is at http://www.examinations.ie/archive/markingschemes/2010/LC020ALP000EV.pdf [Broken] for anyone interested, it's question 6 part a. I'm just questioning their use of the formula for circular motion - surely, the formula only holds if v is a constant, however, the mass is clearly accelerating so this is not the case, so how can they justify using this equation in their solution? Thanks.

at B mgcos(theta)=mv^2/r
 
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  • #13
Thanks for the replies, the bit i don't understand is the use of the centripetal force equation though. In the derivation of the centripetal force f = mv^2/r, it was taken that the mass is moving at a uniform speed and it's acceleration vector was towards the centre of the circle, however if the mass is not moving with uniform speed, then the acceleration vector is not towards the centre of the circle, and in this case how can you use the centripetal force equation? Or can the formula be used no matter whether or not the acceleration/force is pointed towards the centre? Or are we just saying the component of force that is directed towards the centre of the circle is mv^2/r but that's only a component of the entire force? Sorry I'm so confused haha thanks.
 
  • #14
ah ok nevermind i think i get it thanks
 
  • #15
evansmiley said:
Or are we just saying the component of force that is directed towards the centre of the circle is mv^2/r but that's only a component of the entire force?

I think we have a winner
 

1. What is circular motion?

Circular motion is the movement of an object along a curved path, where the object's velocity is constantly changing due to the direction of its motion.

2. What is centripetal force?

Centripetal force is the force that acts on an object moving in circular motion, always directed towards the center of the circle. It is responsible for keeping the object in its circular path.

3. How is centripetal force related to circular motion?

Centripetal force is essential for an object to maintain circular motion. Without this force, the object would move in a straight line tangent to the circle.

4. What factors affect the magnitude of centripetal force?

The magnitude of centripetal force depends on the mass of the object, the speed of the object, and the radius of the circular path. A larger mass or higher speed will result in a larger centripetal force, while a larger radius will result in a smaller centripetal force.

5. What is the difference between centripetal force and centrifugal force?

Centripetal force is the force that keeps an object in circular motion, while centrifugal force is a fictitious force that appears to act on an object moving in a circular path, pushing it away from the center of the circle. Centrifugal force does not actually exist and is a result of the object's inertia.

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