# A tape,pulley,disk and undisclosed mass.

1. May 16, 2007

### MathematicalPhysicist

the question is:
A disk of mass M and radius R unwinds from a tape wrapped around it. the tape passes over a frictionless pulley, and a mass m is suspended from the other end. assume that the disk drops vertically.
1. relate the acclerations of m and the disk, a and A, respectively to the angualr accelration of the disk.
(the answer clue reveals that the naswer is: if A=2a, then alpha=3A/R.
my answer is that the accleration of the disk equals -a+$$\alpha$$R=A and then i get that if A=2a then alpha equals 3a/R, but it's the opposite it should be 3A/R, which i don't see how to arrive at this.

anyway, in the attached file there's a pic of this, the above left pic.

#### Attached Files:

• ###### Scan1.BMP
File size:
68.5 KB
Views:
136
2. May 16, 2007

### chaoseverlasting

I dunno if I'm doing this wrong, but this is what I get:

Let the tension in the string be T, acceleration of the disk be A and of the mass be a and R be the radius of the disk.

For the small mass m:

T-mg=ma ---1

For the disk:

Mg-T=MA

$$TR=I\alpha$$

$$I=\frac{MR^2}{2}$$

$$A=R\alpha$$

Solving these, A=2g/3, T=g/3, $$a=\frac{g(M-3m)}{3m}$$. What did I do wrong?

3. May 16, 2007

### MathematicalPhysicist

the problem is that it's not given to you that the disk rolls without slippering, if it were so, then obviously we would have A=R*(alpha).

4. May 18, 2007

### chaoseverlasting

But if the rope is wound tightly across the disk, then it must roll without slipping as there is no other option. It cant slip (across what?). Therefore a=r(alpha) must hold.

5. May 18, 2007

### chaoseverlasting

In any case, the velocity of the string along the tension must be the same at all the points on the string.

6. May 21, 2007

### MathematicalPhysicist

the problem is that's not a string but a tape!