Angular Velocity Homework: v=r*omega, Answers & Explanation

[louza8]

Homework Statement

http://img690.imageshack.us/img690/5381/screenshot20110430at121.png

Homework Equations

v=r*omega

The Attempt at a Solution

v=0.5m/s
r=0.2m
omega=0.5/0.2
omega=2.5rad/s

(i) Is this correct for the angular velocity of the disk?
(ii) The velocity of the center of the disk? Wouldn't this be equal to the tangential velocity shown on the diagram to be 0.5m/s? The wording of the question is tripping me up, I am not sure of what is being asked of me for this part of the question.

Thanks in advance.

 

[Mike Pemulis]

(i) Is this correct for the angular velocity of the disk?
(ii) The velocity of the center of the disk? Wouldn't this be equal to the tangential velocity shown on the diagram to be 0.5m/s? The wording of the question is tripping me up, I am not sure of what is being asked of me for this part of the question.

(i) Looks good to me.

(ii) Not exactly. The tangential velocity at the top of the disk and the velocity of the center are not the same, so you need to think about how they are related.

Two hints: The disk is rolling without slipping, so what's the speed of the point on the disk touching the ground? If the disk were spinning freely and not traveling forward, what would be the speed of this point?

 

[louza8]

The speed of that point would be -0.5m/s? So is the velocity of the centre point zero?

Edit: is this what is known as the Instantaneous Center of Velocity?

 

[Mike Pemulis]

That would be true if the disk was spinning freely -- it's pretty clear that the two points on opposite sides would move in exactly the opposite direction at the same speed. But that is not true if the disk is not spinning freely. In this situation, the disk is rolling without slipping. So what's the speed of the point on the disk in contact with the ground, at the instant it touches the ground?

 

[louza8]

zero? I don't know, I'm obviously missing points of understanding.

Is the 0.5m/s the velocity as observed from the 'ground' so it is the vector sum of the center of mass' velocity and the tangential component of velocity such that

v_ground=v_center+v_tan=2v_center
0.5=2v
v_cm=0.25m/s?

 

[Mike Pemulis]

zero? I don't know, I'm obviously missing points of understanding.

Is the 0.5m/s the velocity as observed from the 'ground' so it is the vector sum of the center of mass' velocity and the tangential component of velocity such that

v_ground=v_center+v_tan=2v_center
0.5=2v
v_cm=0.25m/s?

The answer is right, but the work not really right -- I think you kind of got lucky. I thought about it and came up with a better way to see the answer (what I said before was valid but more complicated). Let me see if this works better:

The ground is not moving, and the disk is "not slipping" which implies that the point touching the ground is moving at the same velocity of the ground -- which is zero! So the point touching the ground is not moving.

Now draw a line from this point to the top of the disk. This line is tracing out a larger circle, with a radius that is the same as the diameter of the disk. Its end is at the top of the disk, so you know the velocity at the tip of the line -- 0.5 m/s. What's the angular velocity of the line? Is it the same along the entire line? (Yes.)

Stare at this picture. Notice that the center of the disk is also on the line. So what's the tangential velocity of the center?

 

[louza8]

Hey Mike thanks for your help!

I think I grasp it now but using a different explanation (thank you for your time), the following diagram makes sense to me, is this what is going on in this instance?

http://img189.imageshack.us/img189/363/screenshot20110501at114.png

 

[Mike Pemulis]

That definitely works; nice job!