Angular momentum of a disk about an axis parallel to center of mass axis

[vcsharp2003]

Homework Statement

A uniform disk of mass M and radius R is rotating at an angular velocity of $\omega$ about it's center C. What is the angular momentum of the disk about an axis passing through A and perpendicular to the plane of disk.

Relevant Equations

$I_c=\frac {MR^2} {2}$
$\vec L = I \vec {\omega}$

I am using the following formula to solve this problem.
$$ L_a= L_c + \text { (angular momentum of a particle at C of mass M)}$$
Because the point C is at rest relative to point A, so the second term in RHS of above equation is zero. Hence, the angular momentum about A is same as angular momentum about it's center of mass C.

$$\therefore L_a = \frac{MR^2}{2} ~ \omega$$.

I am not sure if above conclusion is correct.

 

[kuruman]

It is correct. You can show formally that the angular momentum of an object about an arbitrary point P is the angular momentum about the object's center of mass plus the angular momentum of the center mass about point P. Here, the latter term is zero as you noted.

 

[vcsharp2003]

It is correct. You can show formally that the angular momentum of an object about an arbitrary point P is the angular momentum about the object's center of mass plus the angular momentum of the center mass about point P. Here, the latter term is zero as you noted.

Since $L = I \omega$, can we say that the angular velocity of the disk about A axis is same as the angular velocity about it's C axis i.e. $\omega_{a} = \omega$?

 

[kuruman]

Since $L = I \omega$, can we say that the angular velocity of the disk about A axis is same as the angular velocity about it's C axis i.e. $\omega_{a} = \omega$?

The angular velocity is the range of change of angle with respect to time. In this case consider angle
$\theta$ formed by a reference line, say along the horizontal and the line from C to a point P on the disk that rotates with it. Then you can say that $\omega_0=\frac{d\theta}{dt}$. Now draw a line from A to P. That line forms angle $\phi$ relative to the horizontal and the angular velocity of P (not of the disk) will be $\omega=\frac{d\phi}{dt}$. You have to show that $\omega=\omega_0$ for any point P on the disk.

 

[vcsharp2003]

The angular velocity is the range of change of angle with respect to time. In this case consider angle
$\theta$ formed by a reference line, say along the horizontal and the line from C to a point P on the disk that rotates with it. Then you can say that $\omega_0=\frac{d\theta}{dt}$. Now draw a line from A to P. That line forms angle $\phi$ relative to the horizontal and the angular velocity of P (not of the disk) will be $\omega=\frac{d\phi}{dt}$. You have to show that $\omega=\omega_0$ for any point P on the disk.

In general, the angle $\theta$ and $\phi$ will not be equal, so the two angular velocities would have different magnitudes.

 

[kuruman]

In general, the angle $\theta$ and $\phi$ will not be equal, so the two angular velocities would have different magnitudes.

Yes. You can see that as the distance between A and C becomes larger, the change of angle $\phi$ with respect to time will become smaller.