Work Energy Theorem and Circular Motion

In summary, when a car is coasting without friction towards a hill with a height of 'h' and a radius of curvature of 'r', the initial speed needed for the car's wheels to lose contact with the road at the top of the hill can be found by setting the kinetic energy at the bottom of the hill equal to the potential energy at the top of the hill. This means that mV^2/r must be greater than mgh, and can also be substituted with g*r. Therefore, the minimum initial speed needed for the car's wheels to lose contact with the road is when a=g.
  • #1
Ilsem
2
0

Homework Statement


A car is coasting without friction toward a hill of height 'h' and radius of curvature 'r'.
What initial speed will result in the car's wheels just losing contact with the roadway as the car crests the hill?


Homework Equations


Kinetic Energy = (1/2)(m)(v^2)
Potential Energy = mgy


The Attempt at a Solution


Because the acceleration will vary with time, constant energy kinematics can't be used to solve the question. Without friction, there are no nonconservative forces acting on the system, so there is no energy lost. Therefore the kinetic energy of the car at the bottom of the hill must be equal(?) to the potential energy of the car at the top of the hill. But I can't seem to work the radius of curvature into the theory at all. There's can't be a centripidal force because the only force holding the car to the curvature of the hill is the force of gravity. I'm a little stuck at this point. Thank you in advance for any help someone can give.
 
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  • #2
Welcome to PF.

When the mv2/r is greater than mg contact force won't that mean that the car will lose contact with the road? So when v2 = g*r.

So your initial mV2 must be greater than m*g*h +mv2/2 where v2 can be substituted with g*r ?
 
  • #3
Thank you very much for the kind welcome.

I see. So if the formula a=(v^2)/r is usually used for centripidal force, and the only force holding the car to the curve is gravity, then this formula has to be used as a minimum where a=g? I think that makes sense.
 

Related to Work Energy Theorem and Circular Motion

1. What is the Work Energy Theorem?

The Work Energy Theorem states that the work done by the net force on an object is equal to the change in kinetic energy of that object.

2. How is the Work Energy Theorem applied in circular motion?

In circular motion, the net force on an object is constantly changing direction, therefore the work done by this force is constantly changing direction as well. This results in a change in the object's velocity, or kinetic energy, even if the speed remains constant.

3. Can the Work Energy Theorem be applied to non-circular motion?

Yes, the Work Energy Theorem can be applied to any type of motion, as long as there is a net force acting on the object. It is a general principle that relates the work done by a force to the change in an object's energy.

4. How is the Work Energy Theorem related to conservation of energy?

The Work Energy Theorem is a manifestation of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. In the case of circular motion, the change in kinetic energy is equal to the work done by the net force, which is equal to the change in potential energy.

5. What are some real-life examples of the Work Energy Theorem in action?

The Work Energy Theorem can be observed in many everyday activities, such as riding a bike, swinging on a playground swing, or throwing a ball. In all of these cases, a force is applied to an object, resulting in a change in its energy. It is also used in understanding the motion of objects in various sports, such as a gymnast on a balance beam or a figure skater performing a routine.

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