# Rigid Body Problem Involving a Tilted Rotating Disc

Peter564
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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L=(1/2)mr^2w

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

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

##\ ##

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why ?
Because of that 45° angle. ##I## will not be ##\frac 12mr^2##

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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?

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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.
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.

• Lnewqban
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There should be a moment about the center of mass, which plane of action rotates with the disc.

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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:
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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}##?

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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.

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