Understanding Kinetic Energy: Moment of Inertia and Rotational Motion

[deuteron]

Homework Statement

Calculate the kinetic energy

Relevant Equations

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Consider the above setup. Here, to get the kinetic energy of the body, the moment of inertia with respect to the $y-$axis has to be calculated. This can be done in two ways:

1. The moment of inertia of the rotation around the center of mass is $\Theta_s$, then the kinetic energy is $T=\frac 12 \Theta_s\dot\varphi^2 + \frac 12 m \dot r_s^2$ where $r_s$ is the location of the center of mass with respect to the coordinate system in the diagramm.

2. The moment of inertia of the rotation around the point $A$ is $\Theta_a$, then the kinetic energy is $T=\frac 12 \Theta_a \dot\varphi^2$.

A quick calculation results in the same kinetic energy for both of the above methods:
$$T= \frac 12 ( \frac 25 -\frac 9{64})mR^2 \dot\varphi^2 +\frac 12 m (R^2\dot\varphi^2 + (\frac 38)^2R^2\dot\varphi^2-2R^2\dot\varphi^2\frac 38 \cos\varphi)$$

What confuses me, is that the point $A$ moves with time, however we don't take its translational motion with respect to the coordinate frame into account, why?

 

[Lnewqban]

What is the meaning of the represented theta angle in the upper diagram?

The distance between the location of the CM and the periphery of the semi sphere increases as the phi angle increases.

The kinetic energy of the semi sphere when its CM reaches the lowest spatial point should equal the difference in potential energy experimented by the CM when relocated away from the balance point.
Perhaps you could verify your results comparing both types of energy.

 

[deuteron]

What is the meaning of the represented theta angle in the upper diagram?

The distance between the location of the CM and the periphery of the semi sphere increases as the phi angle increases.

The kinetic energy of the semi sphere when its CM reaches the lowest spatial point should equal the difference in potential energy experimented by the CM when relocated away from the balance point.
Perhaps you could verify your results comparing both types of energy.

Both methods yield the same result (both methods: translational kinetic energy of the center of mass + rotational kinetic energy with respect to the center of mass = rotational kinetic energy with respect to the point $A$), but what I don't understand is why we don't take the translational kinetic energy of the point $A$ into account.

 

[Steve4Physics]

... but what I don't understand is why we don't take the translational kinetic energy of the point $A$ into account.

When rolling without slipping, the instantaneous velocity of the object at point A is zero.

That means during any sufficiently small time-interval, the object is behaving the same as if it were rotating about a fixed pivot at A. So the kinetic energy at any instant can be found by considering only the rotation about A.

 

[deuteron]

When rolling without slipping, the instantaneous velocity of the object at point A is zero.

That means during any sufficiently small time-interval, the object is behaving the same as if it were rotating about a fixed pivot at A. So the kinetic energy at any instant can be found by considering only the rotation about A.

Thanks! So only instantaneous motion is important, the overall motion is not, when calculating the kinetic energy, have I understood right?

 

[Steve4Physics]

Thanks! So only instantaneous motion is important, the overall motion is not, when calculating the kinetic energy, have I understood right?

You're welcome.

But I don’t think that’s right (though I may have misunderstood). In this problem we want to find the overall (total) instantaneous kinetic energy ($K_{overall}$). This is the sum of the instantaneous translational kinetic energy ($K_{trans}$) and the instantaneous rotational kinetic energy ($K_{rot}$).

Using a different axis of rotation gives a different split between $K_{trans}$ and $K_{rot}$ - but their total, $K_{overall}$, is unaffected by the choice of axis.

I think the key is to understand the geometry/kinematics. Consider the ‘bit’ of the object in instantaneous contact with point A (not point A itself, which is moving). The instantaneous velocity of this ‘bit’ is zero so we can attribute all of the instantaneous overall kinetic energy to rotation about A.

I imagine a formal/rigorous proof is possible by considering motion during some time interval $\delta t$ and taking the limit as $\delta t \rightarrow 0$. But I’m not going to try! Try this link for a discussion.