# Rotational Kinetic Energy and Conservation of Momentum

• aliens123
In summary, the difference in rotational inertia between a solid sphere and a hoop rolling down an inclined plane results in different net forces and acceleration, due to the friction force acting on the objects at the point of contact with the incline. Additionally, for a solid sphere and a hoop with similar materials, the static friction force will not be the same, as it is determined by the coefficient of friction which only gives the maximum force possible. The actual force is determined by the need to prevent slippage. For a rod experiencing an impulse at the center of mass or at the top, the kinetic energy will be greater for the rod pushed at the top due to the force acting over a greater distance.
aliens123
I am having trouble wrapping my head around a physics concept.

If we roll solid sphere down an inclined plane it will reach the bottom at a different time than if we were to say, roll a hoop down the same inclined plane. This is because they have different rotational inertias, and so more of the potential energy will go into rotational energy.

But doesn't this conflict with the law that the acceleration of the center of mass of a system of particles depends only on the net force acting on the system of particles, and therefore the center of mass of a solid sphere and a hoop should reach the bottom of the inclined plane at the same time?

Another thing that is confusing me. In a physics class we were shown an example where a rod experienced an impulse either at the center of mass of the rod, or at the top end of the rod. The question was which will have more kinetic energy.

The answer was that because the acceleration of the center of mass was the same for both, and the rod pushed at the top also had some rotational energy, the kinetic energy of the rod pushed at the top (and not directly in the center) was greater. The extra energy came from the fact that although the impulse was the same, the work done was not the same because the rod which was pushed from the top had its force act over a greater distance.

But doesn't this above scenario seem to conflict with the first scenario presented?

Would appreciate any help,
Thank You.

There is also a (static) friction force acting on the rolling objects at the point of contact with the incline -- as there must be, in order to produce a toque about the object's center and cause it to rotate as it rolls.

Anyway, that friction force is different for the two objects of different rotational inertia, so the net force and acceleration are different as well.

aliens123
Redbelly98 said:

There is also a (static) friction force acting on the rolling objects at the point of contact with the incline -- as there must be, in order to produce a toque about the object's center and cause it to rotate as it rolls.

Anyway, that friction force is different for the two objects of different rotational inertia, so the net force and acceleration are different as well.

If we held the surface area to be the same, why would the frictional force be different? Furthermore, if we assume that the frictional force is only the simplified version of the normal force and the coefficient of friction, then with similar materials the frictional force should be the same.

aliens123 said:
if we assume that the frictional force is only the simplified version of the normal force and the coefficient of friction, then with similar materials the frictional force should be the same.
No, this is not how static friction works. Normal force times static friction coefficient gives you the maximal static friction force possible, not the actual force acting.

A.T. said:
No, this is not how static friction works. Normal force times static friction coefficient gives you the maximal static friction force possible, not the actual force acting.

I am not saying that the static frictional force is proportional to the coefficient. The coefficient merely gives a maximal static friction force possible, like you said. But this still does not explain why the static frictional force will be different for the two rolling objects.

aliens123 said:
But this still does not explain why the static frictional force will be different for the two rolling objects.
Static frictional force will take whatever value is needed to prevent slippage (within the bounds given by the coefficient).

aliens123
The friction force times the radius of the hoop or sphere equals the torque applied to the hoop or sphere. The torque in turn is related to the acceleration in the direction of the plane. I did this recently, but now search doesn't show my last post on this. So redoing the math again:

force in the direction of the plane related to gravity = fg = m g sin(θ)
force in the direction of the plane related to friction = ff= α I / r, where α is angular acceleration, I is angular inertia, r is radius
linear acceleration in the direction of the plane = a = α r = (fg - ff)/m = g sin(θ) - a I/(m r^2)
a(1 + I/(m r^2)) = g sin(θ)
a = g sin(θ) / (1 + I/(m r^2))
ff = m g sin(θ) - m a

for hoop or hollow cylinder I = m r^2, a = 1/2 g sin(θ), ff = 2/4 m g sin(θ)
for hollow sphere I = 2/3 m r^2, a = 3/5 g sin(θ), ff = 2/5 m g sin(θ)
for solid cylinder I = 1/2 m r^2, a = 2/3 g sin(θ), ff = 2/6 m g sin(θ)
for solid sphere I = 2/5 m r^2, a = 5/7 g sin(θ), ff = 2/7 m g sin(θ)

As for the rod, the linear acceleration equals the linear force / mass, regardless of where the force is applied, but if the force is applied at the end of the rod, then the rate of acceleration of the end of the rod is greater than the rate of acceleration of the center of mass of the rod, and the force is applied over a greater distance.

aliens123
aliens123 said:
The coefficient merely gives a maximal static friction force possible, like you said. But this still does not explain why the static frictional force will be different for the two rolling objects
So if the coefficient only determines the maximum force, then what determines the actual force? When you work one of these problems, what is the process you use to determine it?

aliens123

## 1. What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotational motion. It is dependent on the object's moment of inertia and angular velocity.

## 2. How is rotational kinetic energy calculated?

The formula for rotational kinetic energy is K = 1/2 * I * ω², where K is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

## 3. What is conservation of momentum?

Conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant over time. This means that the total momentum before and after a collision or interaction remains the same.

## 4. How is conservation of momentum related to rotational kinetic energy?

Conservation of momentum applies to rotational motion just as it does to linear motion. In a closed system, the total angular momentum before and after a collision or interaction remains the same. This means that the sum of the initial rotational kinetic energy and the initial potential energy is equal to the sum of the final rotational kinetic energy and final potential energy.

## 5. Can rotational kinetic energy be converted into other forms of energy?

Yes, rotational kinetic energy can be converted into other forms of energy, such as heat or sound. This can occur through friction or other forces acting on the rotating object, causing it to slow down and lose kinetic energy.

• Mechanics
Replies
19
Views
1K
• Mechanics
Replies
3
Views
1K
• Mechanics
Replies
3
Views
1K
• Mechanics
Replies
2
Views
971
• Mechanics
Replies
3
Views
2K
• Mechanics
Replies
3
Views
1K
• Mechanics
Replies
2
Views
245
• Mechanics
Replies
2
Views
918
• Mechanics
Replies
4
Views
215
• Mechanics
Replies
5
Views
1K