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

In summary: As a result, the angular velocity ##\omega## will be smaller than ##\omega_{a}## In summary, the formula for solving the problem involving angular momentum is given by $$ L_a= L_c + \text { (angular momentum of a particle at C of mass M)}$$ Because the point C is at rest relative to point A, the second term in the right-hand side of the equation is zero. This means that the angular momentum about A is the same as the angular momentum about its center of mass C. Additionally, the angular velocity of the disk about A axis is not the same as the angular velocity about its C axis, as the angle between the two axes will generally not be equal. As the distance between
  • #1
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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.

IMG_20220601_175235__01.jpg
 
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  • #2
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.
 
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  • #3
kuruman said:
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##?
 
  • #4
vcsharp2003 said:
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.
 
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  • #5
kuruman said:
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.
 
  • #6
vcsharp2003 said:
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.
 
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