Rigid Body Problem Involving a Tilted Rotating Disc

In summary: So, I'll give your method a try. Thanks.In summary, the conversation discusses the calculation of angular momentum and kinetic energy for a rotating disk using different equations. The correct method is to calculate the moment of inertia tensor and project it onto the rotation axis to obtain the effective moment of inertia. This approach may be more advanced for introductory physics and an alternative method of calculating the total angular momentum as the sum of two components is suggested.
  • #1
Peter564
2
0
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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  • #2
:welcome: !

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 ?

##\ ##
 
  • #3
BvU said:
why ?
Because of that 45° angle. ##I## will not be ##\frac 12mr^2##
 
  • #4
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?
 
  • #5
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.
 
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  • #6
There should be a moment about the center of mass, which plane of action rotates with the disc.
 
  • #7
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:
  • #8
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}##?
 
  • #9
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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1. What is a rigid body problem involving a tilted rotating disc?

A rigid body problem involving a tilted rotating disc is a physics problem that involves analyzing the motion of a disc that is both rotating and tilted at an angle. This type of problem is often used to study the dynamics of rotating objects and can be applied to various real-world scenarios, such as the motion of a spinning top or a gyroscope.

2. What are the key factors that affect the motion of a tilted rotating disc?

The key factors that affect the motion of a tilted rotating disc include the initial angular velocity of the disc, the angle of tilt, the mass and shape of the disc, and any external forces acting on the disc, such as friction or air resistance.

3. How is the angular velocity of a tilted rotating disc calculated?

The angular velocity of a tilted rotating disc can be calculated using the formula ω = v/r, where ω is the angular velocity, v is the tangential velocity of a point on the disc, and r is the distance from the center of the disc to that point. This formula can be applied to any point on the disc, as long as the disc is rigid and not deforming.

4. What is the significance of the moment of inertia in a rigid body problem involving a tilted rotating disc?

The moment of inertia is a measure of an object's resistance to rotational motion. In a rigid body problem involving a tilted rotating disc, the moment of inertia plays a crucial role in determining the disc's angular acceleration and the distribution of mass within the disc affects the moment of inertia. A larger moment of inertia means the disc will be more resistant to changes in rotational motion.

5. How does the angle of tilt affect the motion of a tilted rotating disc?

The angle of tilt can affect the motion of a tilted rotating disc in several ways. As the angle of tilt increases, the disc's angular velocity may decrease due to increased friction and air resistance. Additionally, the disc's center of mass may shift, causing changes in the moment of inertia and altering the disc's motion. The angle of tilt can also affect the stability of the disc, with higher angles of tilt making it more likely for the disc to topple or wobble.

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