# What is the effect of a tangential force on a rigid body in terms of kinetic energy? [duplicate]

Let's take into consideration a sphere. We apply a force F tangent to the sphere.
We know that the linear acceleration of that sphere will be equal to F/m where m is the total mass of the sphere.

Then calculating the total kinetic energy, we see that we have the translational kinetic energy + rotational kinetic energy (since the force is not applied on the center of mass).

Now, if we had applied the same force on the center of mass we would have a kinetic energy which depends only on the translational motion of the body since there wouldn't be any torque involved. This energy is lower than the one considered above (when the force was not applied on the center of mass). It seems that if we apply the force tangential to the sphere we have more kinetic energy out of nothing. Why?

Every thinking person asks this question or the equivalent when learning this stuff. So it’s prolly a good sign.

Because the same amount of force applied over a larger distance is more work energy $$\int Fdx$$ added to the system. The case of applying the force off-center means the force gets applied over more distance and hence does more work and hence adds more energy.

It seems that if we apply the force tangential to the sphere we have more kinetic energy out of nothing. Why?

You are not getting more kinetic energy out of nothing. The additional energy is supplied by the source of the force. Kinetic energy is the energy of motion. Force is not energy.

Force transfers energy from some source to the sphere when it does work. Conservation of energy dictates that the source of a force that is applied tangentially to the sphere transfers more energy (does more work of $$W=\tau \theta+Fd$$) than the source of the force that is only applied to the center of mass of the sphere (which does in less work $$W=Fd$$).

Hope this helps.

This is a really good question ! Although Al Brown's answer is really good, I wanted to elaborate a bit further.

What is completely general is the Work-Energy Theorem which states that ∫𝐹𝑑𝑥 is equal to change in kinetic energy (this may be rotational or translational). In your question, you implicitly assumed that the friction doesn't exist. Without friction, we can write down the Newton's Equations of Motion for this particle (for sum of torques and forces):

F = ma (for linear motion) and FR = Ia/R (for rotational motion),

where I is the moment of inertia about the centre and R is the radius (Note that I have assumed the rotational acceleration to be equal to a/R which is not quite right as will be explained shortly). The second equation implies,

F = Ia/Rˆ2.

However, this is in contradiction with the first equation. I is equal to some factor times m*Rˆ2, so according to our second equation F = (some factor)*ma which may not always be right.

This happened because we have assumed that the ball was not slipping. This is not true because there is no friction force to balance this out ! This means that contrary to rolling without slipping, the point where the ball touches the ground is not stationary (in the case for rolling without slipping, the backwards angular motion and the forwards linear motion cancel at this point) and hence the ball will move more in a given amount of time due to the additional angular motion.

However, if there is friction and it is rolling without slipping, the ball will actually travel the same distance as a mass which can't roll. However, in this case, the work done by the force will be shared by the rotational kinetic energy and translational kinetic energy. A good example is the fact that in an inclined plane a full cylinder, sphere and a hollow cylinder doesn't arrive to the bottom at the same time.

• Good answer! For the whole details we had fin doing this: physics.stackexchange.com/a/659029/307354 Aug 20, 2021 at 19:50
• Thank you very much ! Aug 21, 2021 at 8:30