Rigid Body Problem Involving a Tilted Rotating Disc

AI Thread Summary
The discussion centers on calculating the angular momentum and kinetic energy of a tilted rotating disc. Initial calculations yielded L=0.02 Nms, I=0.02 Kgm², and KE=10 mJ, but participants questioned the appropriateness of the method due to the 45° tilt. The moment of inertia must account for the tilt, requiring the use of a moment of inertia tensor to accurately determine angular momentum. Participants suggest that the total angular momentum can be derived from the components along the perpendicular and parallel axes of rotation. The conversation emphasizes the need for a more advanced understanding of rotational dynamics to solve the problem correctly.
Peter564
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Homework Statement
A uniform disc of radius 0.1 m and mass 0.4 kg is rotating with angular velocity 1 rad s−1 about
an axis at 45◦ to its plane through its centre of mass. What is
(a) its angular momentum, and (b) its kinetic energy?
(You may assume the centre of mass is stationary.)
Relevant Equations
L=(1/2)mr^2w KE=(1/2)Iw^2
I=(1/2)mr^2
Using these equations, I find L=0.02Nms, I=0.02Kgm^2 and KE=10mJ

However, i don't think that this is the right method here.
 
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Peter564 said:
L=(1/2)mr^2w

Using these equations, I find L=0.02Nms
Using these equations I get something else. Check your calculations ...

Peter564 said:
don't think that this is the right method here
why ?

##\ ##
 
BvU said:
why ?
Because of that 45° angle. ##I## will not be ##\frac 12mr^2##
 
Peter564 said:
What is
(a) its angular momentum
Are you supposed to find the angular momentum vector ## \vec L## relative to an origin at the center of the disk?

Or, are you supposed to find just the component of ##\vec L## along the axis of rotation?
 
Peter564 said:
Homework Statement:: A uniform disc of radius 0.1 m and mass 0.4 kg is rotating with angular velocity 1 rad s−1 about
an axis at 45◦ to its plane through its centre of mass. What is
(a) its angular momentum, and (b) its kinetic energy?
(You may assume the centre of mass is stationary.)
Relevant Equations:: L=(1/2)mr^2w KE=(1/2)Iw^2
I=(1/2)mr^2

Using these equations, I find L=0.02Nms, I=0.02Kgm^2 and KE=10mJ

However, i don't think that this is the right method here.
It is not the right method. The moment of inertia ##I=\frac{1}{2}mR^2## is about an axis that is perpendicular to the disk and passes through its center. The axis here is not perpendicular to the plane of disk but inclined at 45° from it.
 
There should be a moment about the center of mass, which plane of action rotates with the disc.
 
If the OP question is a test after learning about the moment of inertia tensor, I will suggest writing up that matrix using principal coordinates and project it onto the rotation axis expressed in that coordinate system to get the effective moment of inertia around that axis. That is the general approach anyway.

Edit: removed unnecessary step about rotating before projection.
 
Last edited:
Angular momentum is a vector.
If the disk is rotating about an an axis perpendicular to the plane of the disk, the angular momentum vector is ##~\mathbf{L}_{\perp}=I_{\perp}{\omega~}\mathbf{\hat e}_{\perp}.##
If the disk is rotating about an an axis parallel to the plane of the disk, the angular momentum vector is ##~\mathbf{L}_{\parallel}=I_{\parallel}{\omega~}\mathbf{\hat e}_{\parallel}.##
What is ##~\mathbf{L}_{\text{tot}}=\mathbf{L}_{\perp}+\mathbf{L}_{\parallel}##?
 
kuruman said:
What is ##~\mathbf{L}_{\text{tot}}=\mathbf{L}_{\perp}+\mathbf{L}_{\parallel}##?
If ##\omega## is the angular speed around the tilted axis then surely this yields too large a result?

I would just write it as ##\mathbf{L} = \mathbf{n}^T \, \mathbf{I} \, \mathbf{n} \, \omega##, where ##\mathbf{n}## is the rotation unit axis and ##\mathbf{I}## is the 3x3 moment of inertia, which is very simple to write up if principal coordinates are selected.
 
  • #10
Filip Larsen said:
If ##\omega## is the angular speed around the tilted axis then surely this yields too large a result?
I am not sure about that if the relation between your ##\mathbf{\hat n}## and my ##\mathbf{\hat e}_i## is something like ##\mathbf{\hat n}=\mathbf{\hat e}_{\parallel}\cos\!\varphi+\mathbf{\hat e}_{\perp}\sin\!\varphi##. Your approach is certainly simple but probably above the level of introductory physics. The disk here is rotating about two perpendicular axes with the same angular speed. Therefore, it should be easy to explain to someone who has seen vectors but not tensors that the total angular momentum can be viewed as the resultant of two components that can be calculated separately.
 
  • #11
kuruman said:
I am not sure about that if the relation between your ##\mathbf{\hat n}## and my ##\mathbf{\hat e}_i## is something like ##\mathbf{\hat n}=\mathbf{\hat e}_{\parallel}\cos\!\varphi+\mathbf{\hat e}_{\perp}\sin\!\varphi##.
Yes, that is my notion of the effective rotation axis. The two approaches do seem to equal each other if the ##\cos\varphi## and ##\sin\varphi## factors are included in the parallel and perpendicular angular momentum vectors since this then corresponds to ##\omega## projected onto the parallel and perpendicular axis, respectively.
 
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