Velocity & Accel of a mass inside a slot on a rolling disk

AI Thread Summary
The discussion focuses on calculating the velocity and acceleration of a mass moving inside a slot on a rolling disk with constant angular velocity. The initial velocity of the disk is established as v_A = RΩ, where R is the radius and Ω is the angular velocity. The challenge arises in determining the velocity of the mass relative to the disk, with attempts to relate it to angular displacement θ using trigonometric functions. The concept of the instantaneous center of rotation is introduced to aid in understanding the changing dynamics as the disk rolls. Overall, the participants are working through the complexities of relating the mass's motion to the disk's rolling motion.
zealeth
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Homework Statement


The disk in Figure 3.30 of radius R rolls without slipping with constant angular velocity Ω. Carved inside the disk is a slot and a mass moves inside the slot. Denoting the position of the mass inside the slot by s, calculate the velocity and acceleration of the mass as a function of θ.

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Homework Equations


[/B]
v_A = RΩ
v_B = v_A + v_B/A
a = dv/dt

The Attempt at a Solution



v_A = R*Ω, which is fairly obvious since the disk does not slip. v_B is then = R*Ω + v_B/A
Finding v_B/A is where I'm having issues. My first thought is that it would be Ω*√((5R/8)^2+s^2), but this isn't in terms of θ and is only angular velocity. I've thought about trying to relate θ using the relation tan(θ) = 8s/(5R), but I don't know how to fit this into the equation.

I haven't taken a look at finding the acceleration yet, but I imagine I could just take the time derivative of the velocity equation to get the acceleration.
 
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It might help to think in terms of instantaneous centre of rotation. Where is that for a rolling disc?
 
haruspex said:
It might help to think in terms of instantaneous centre of rotation. Where is that for a rolling disc?

The instantaneous center is at the point where it contacts the ground, or in this case, point C. I don't know how I would relate this to the velocity of the mass since the mass's instant center is changing as the disk rolls.
 
zealeth said:
The instantaneous center is at the point where it contacts the ground, or in this case, point C. I don't know how I would relate this to the velocity of the mass since the mass's instant center is changing as the disk rolls.
Sure, there is a velocity relative to the disk as well, but the direction of that is constrained. So it remains to find the total speed. How might you do that?
 
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