# Kinetic energy of a rolling disc when slipping

To derive kinetic energy expression for a rolling body, we consider the instantaneous center of rotation and write the angular energy about that point. After some algebraic simplification, we conclude at:

$$K =\frac{1}{2} (I_C + MR^2) \omega^2 \tag{1}$$

Where $$I_c$$ is the inertia of the body about the center of mass, $$MR^2$$ is the extra inertia term from the parallel axis theorem ( shift from the ICAOR to the center of mass), and $$\omega$$ is the angular velocity at which the object rotates around and V is the velocity of the center of mass.

Now, if we had a body that was slipping then would the same line of proof/reasoning hold for writing its kinetic energy?

The reason I ask is that when the body is slipping the rotational speed about com axis is greater than the kinetic component, I don't think the energy can be fully expressed in terms of angular velocity about ICAOR. Or does angular velocity about ICAOR change accordingly to account for this?

The derivation of (1) can be found in open stax

No. The kinetic energy of a rotating body can be expressed as a sum of translational and rotational parts which are generically uncoupled from one another. The no-slip condition provides a relationship between the center of mass velocity and angular velocity, allowing you to express the kinetic energy completely in terms of one or the other. In the absence of such a condition, you need both.

It can be shown that the total kinetic energy of a rigid body can be written

$$T = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$

where $$v_{cm}$$ is the velocity of the center of mass and $$I_{cm}$$ is the moment of inertia about the axis parallel to $$\boldsymbol \omega$$ and passing through the center of mass.

If we have a relationship between $$v_{cm}$$ and $$\omega$$, we can express one in terms of the other. For example, if $$v_{cm} = \beta \omega$$ for some constant $$\beta$$, then this becomes

$$T = \frac{1}{2}m\beta^2 \omega^2 + \frac{1}{2}I_{cm}\omega^2 = \frac{1}{2}\big(m\beta^2 + I_{cm}\big)\omega^2$$

For a disk in pure roll, we have that $$\beta = R$$ (the radius of the disk), which reproduces the result in your link. If the disk is not in pure roll, the best we can do is write

$$T = \frac{1}{2}\big( \frac{mv_{cm}^2}{\omega^2} + I_{cm}\big) \omega^2$$

which depends on the (possibly variable) ratio $$v_{cm}/\omega$$.

• @Buraian In hindsight, I don't think my comments were helpful or entirely correct, so I've deleted them and amended my answer to explain further. Oct 23, 2020 at 21:55

let me tell you an amazing thing.

while pure rolling (when the body is not slipping) You use the concept of ICOR(instantaneous center of rotation) because the point of disc in contact with the ground is at rest and necessary condition to study a rotating body is that the axis should be fixed. Now the axis is moving with time hence for some other frame of reference the axis might be fixed and the the energy you are calculating is the energy of disk in that frame. So you have to consider the

rotational energy of body in that frame + kinetic energy of the frame