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.

 

[BvU]

!

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 ?

$\$

 

[haruspex]

why ?

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

 

[TSny]

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?

 

[kuruman]

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]

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

 

[Filip Larsen]

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.

 

[kuruman]

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 $~\mathbf{L}_{\text{tot}}=\mathbf{L}_{\perp}+\mathbf{L}_{\parallel}$?

 

[Filip Larsen]

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.

 

[kuruman]

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.

 

[Filip Larsen]

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.