A massless disk with an embedded particle rolls down an inclined plane

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  • #1
annamal
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Homework Statement
A massless disk of radius R has an embedded particle
of mass m at a distance R/2 from the center. The disk is released from rest in the position
shown below and rolls without slipping down the fixed inclined plane. Find the velocity
Relevant Equations
v = dr/dt
This is the problem:
Screenshot 2024-03-13 at 4.30.43 PM.png

I am a little confused at the solution below. Since ##\vec r_{P/C} = (R/2)*(5 + 4cos(\theta))^{0.5}\ \vec e_r##, I am wondering why there is no vector e_r component for the velocity. I know that d(R)/dt = 0, but you can still get a derivative in the ##\vec e_r## direction from dcos(theta)/dt

Screenshot 2024-03-13 at 4.31.04 PM.png
 
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  • #2
The solution is unclear to me. In the ground frame, ##\dot{\hat e_r}## is nonzero, but ##{\hat e_r}## is treated as a constant.
Much easier to note that using ##\hat i## parallel to the slope, etc., the velocity of O is ##R\dot\theta\hat i## and relative to that the velocity of P is ##\frac 12R\dot\theta(\cos(\theta)\hat i-\sin(\theta)\hat j)##. Add those and sum the squares of the orthogonal components.
 
  • #3
The unit vectors ##\hat e_r## and ##\hat e_{\theta}## are not drawn very accurately in the figure here:
1710385216827.png


##\hat e_r## should be in the direction of ##\vec r^{P/C}## as shown here:
1710385242229.png

[EDIT: I'm not sure about the unit vector ##\hat e_{\theta}##. I drew it as being perpendicular to ##\hat e_r##. But, maybe ##\hat e_{\theta}## is defined to be perpendicular to line OP. If so, then ##\hat e_{\theta}## is not perpendicular to ##\hat e_r##. Oh, wait. In the solutions it states ##\hat k = \hat e_r \times \hat e_{\theta}##, so these three unit vectors are mutually perpendicular, and the direction of ##\vec v^{P/C}## is ##\hat e_{\theta}##.]



haruspex said:
In the ground frame, ##\dot{\hat e_r}## is nonzero, but ##{\hat e_r}## is treated as a constant.
I don't think they treated ##\hat e_r## as a constant. They didn't find ##\dot {\vec r}^{P/C}## by taking the time derivative of ##\vec r^{P/C}##.

They used the fact that the velocity of ##P## relative to the lab is the same as the velocity of P relative to the instantaneous center of rotation of the disk (at C). Thus, ##\dot {\vec r}^{P/C} = \vec{\omega}^{\rm disk} \times \vec r^{P/C}##, where ##\vec{\omega}^{\rm disk} = \dot {\theta} \hat k##.
 
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  • #4
TSny said:
The unit vectors ##\hat e_r## and ##\hat e_{\theta}## are not drawn very accurately in the figure here:
View attachment 341760

##\hat e_r## should be in the direction of ##\vec r^{P/C}## as shown here:
View attachment 341761



I don't think they treated ##\hat e_r## as a constant. They didn't find ##\dot {\vec r}^{P/C}## by taking the time derivative of ##\vec r^{P/C}##.

They used the fact that the velocity of ##P## relative to the lab is the same as the velocity of P relative to the instantaneous center of rotation of the disk (at C). Thus, ##\dot {\vec r}^{P/C} = \vec{\omega}^{\rm disk} \times \vec r^{P/C}##, where ##\vec{\omega}^{\rm disk} = \dot {\theta} \hat k##.
Ok.
Do you know what that LHS superscript notation means?
##^{R^?}v^{P/C}##
It is too blurry to read the "?".
 
  • #5
haruspex said:
Do you know what that LHS superscript notation means?
##^{R^?}v^{P/C}##
It is too blurry to read the "?".
It looks like ##R'##. This notation ##R'## appears in the first figure near the top of the ramp. Maybe it's something to do with denoting the frame of reference. Thus, ##^{R\,'} \vec v^{P/C}## is the velocity of P relative to C in the frame ##R'## as opposed to the frame of the rotating disk. I'm not sure.
 
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  • #6
TSny said:
It looks like ##R'##. This notation ##R'## appears in the first figure near the top of the ramp. Maybe it's something to do with denoting the frame of reference. Thus, ##^{R\,'} \vec v^{P/C}## is the velocity of P relative to C in the frame ##R'## as opposed to the frame of the rotating disk. I'm not sure.
Yes it is ##^{R\,'} \vec v^{P/C}##
 
  • #7
TSny said:
It looks like ##R'##. This notation ##R'## appears in the first figure near the top of the ramp. Maybe it's something to do with denoting the frame of reference. Thus, ##^{R\,'} \vec v^{P/C}## is the velocity of P relative to C in the frame ##R'## as opposed to the frame of the rotating disk. I'm not sure.
It is ##^{R\,'} \vec v^{P/C}## and means the reference frame of R' which is the reference frame of the slope and disk.
 
  • #8
haruspex said:
Ok.
Do you know what that LHS superscript notation means?
##^{R^?}v^{P/C}##
It is too blurry to read the "?".
It is ##^{R\,'} \vec v^{P/C}## and means the reference frame of R' which is the reference frame of the slope and disk.
 
  • #9
annamal said:
am wondering why there is no vector e_r component for the velocity
As @TSny wrote, ##\hat e_r## is always along the line CP, and ##\hat e_\theta## is always normal to that. Since C is the instantaneous centre of rotation, there is no velocity component along CP.
 
  • #10
haruspex said:
As @TSny wrote, ##\hat e_r## is always along the line CP, and ##\hat e_\theta## is always normal to that. Since C is the instantaneous centre of rotation, there is no velocity component along CP.
ok but how do you explain this mathematically? I thought ##\frac{d\vec r}{dt} = \vec v^{P/C}##.
Since ##\vec r^{P/C} = (R/2)*(5 + 4 cos\theta)^{1/2} \vec e_r##,
the derivative of that is
##(R/4)*(5 + 4cos\theta)(-4sin\theta)^{-1/2}\dot{\theta} \vec e_r + ...\vec e_{\theta}##. Mathematically like that, the only way to have no ##\vec e_r## is for ##\dot{\theta} = 0##
 
  • #11
annamal said:
ok but how do you explain this mathematically? I thought ##\frac{d\vec r}{dt} = \vec v^{P/C}##.
Since ##\vec r^{P/C} = (R/2)*(5 + 4 cos\theta)^{1/2} \vec e_r##,
the derivative of that is
##(R/4)*(5 + 4cos\theta)(-4sin\theta)^{-1/2}\dot{\theta} \vec e_r + ...\vec e_{\theta}##. Mathematically like that, the only way to have no ##\vec e_r## is for ##\dot{\theta} = 0##
The confusion arises from the definition of C.
For the logic used in the text, and as described by @TSny , C is a point fixed on the ramp where contact is made at some instant. With that definition, the equation for ##\vec r^{P/C}## does not remain true over time.
If we change it to be defined as the point of contact at each instant then the equation can be validly differentiated, but then the derivative does have an ##\vec e_r## term.
 
  • #12
Note that kinetic energy is invariant under rotation of the system. The kinetic energy as a function of theta, therefore, is the same as for the mass rolling along a flat surface. Taking the origin at the centre of the mass we have:
$$x = R\theta + \frac R 2 \sin \theta, \ y = \frac R 2 \cos \theta$$$$\dot x = R\dot \theta(1 + \frac 1 2 \cos \theta), \ \dot y = -\frac 1 2 R\dot \theta \sin \theta$$$$T = \frac 1 2 m(\dot x^2 + \dot y^2) = \frac 1 2 mR^2\dot \theta^2\big(\frac 5 4 + \cos \theta \big )$$$$= \frac {mR^2}{8}\big(5 + 4\cos \theta \big )\dot \theta^2$$
 
  • #13
PeroK said:
Note that kinetic energy is invariant under rotation of the system. The kinetic energy as a function of theta, therefore, is the same as for the mass rolling along a flat surface.
Could you explain that a little further?
I am probably confused, but I visualize a cycloid trajectory (respect to the sloped surface) of the embedded particle P of mass m.

Induced solely by gravity acting on that mass, I see the rotational velocity of the disk as not constant but increasing while reaching a point prior to C (vertical line with the center of the disk) and decreasing while moving away from that point of minimum potential energy respect to C.

1*FweOU8YU3U3CvsDWHNVFNA.png
 
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  • #14
Lnewqban said:
Could you explain that a little further?
Clearly the function ##T(\theta, \dot \theta)## is independent of the angle of the slope.
 
  • #15
PeroK said:
Clearly the function ##T(\theta, \dot \theta)## is independent of the angle of the slope.
Yes, better would have been "theta and its time derivatives".
 
  • #16
haruspex said:
Yes, better would have been "theta and its time derivatives".
I must have been in a Lagrangian mood.
 
  • #17
TSny said:
They used the fact that the velocity of ##P## relative to the lab is the same as the velocity of P relative to the instantaneous center of rotation of the disk (at C). Thus, ##\dot {\vec r}^{P/C} = \vec{\omega}^{\rm disk} \times \vec r^{P/C}##, where ##\vec{\omega}^{\rm disk} = \dot {\theta} \hat k##.
What is this "lab" you are referring to?
 
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  • #18
annamal said:
What is this "lab" you are referring to?
@TSny means the "lab frame". Also known as the ground frame.
 

FAQ: A massless disk with an embedded particle rolls down an inclined plane

What is a massless disk?

A massless disk is a theoretical concept used in physics to simplify problems. It refers to a disk that has no mass of its own, meaning it does not contribute to the system's inertia or gravitational force. In this scenario, the disk's mass is considered negligible compared to the embedded particle.

How does the embedded particle affect the motion of the disk?

The embedded particle, having mass, introduces inertia and gravitational force to the system. As the disk rolls down the inclined plane, the particle's mass influences the acceleration and the rotational dynamics of the disk. The particle's position relative to the center of the disk can also affect the system's behavior.

What forces act on the system as it rolls down the inclined plane?

The main forces acting on the system include gravitational force, normal force from the inclined plane, and frictional force. The gravitational force causes the system to accelerate down the plane, the normal force acts perpendicular to the plane, and the frictional force prevents slipping and allows rolling motion.

How is the acceleration of the system determined?

The acceleration of the system can be determined using Newton's second law and the rotational dynamics equations. By analyzing the forces and torques acting on the disk-particle system, one can derive the linear acceleration of the center of mass and the angular acceleration of the disk.

What assumptions are made in this problem?

Several assumptions are typically made to simplify the analysis: the disk is massless, the particle is rigidly embedded in the disk, the inclined plane is frictional enough to prevent slipping, and air resistance is negligible. These assumptions allow for a more straightforward application of physical principles to solve the problem.

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