# Two wheels and angular acceleration

1. May 17, 2008

### rosstheboss23

[SOLVED] Two wheels and angular acceleration

1. The problem statement, all variables and given/known data
In the figure below, wheel A of radius rA = 11 cm is coupled by belt B to wheel C of radius rC = 25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2. Find the time needed for wheel C to reach an angular speed of 120 rev/min, assuming the belt does not slip. Linear speeds of two rims must be equal.

2. Relevant equations
v=wr; a= w(squared)r

3. The attempt at a solution
v=wr so v1=v2? Then I tried to use this equation w(squared)r=a And then tried using the angular acceleration equations(that are like the kineticmatics equations). The answers I got using these equations though I think should be wrong because the acceleration is give above. Can someone help explain the concept. I can do the brute work if I know.

2. May 17, 2008

### Hootenanny

Staff Emeritus
Hey ross,

I'm not quite sure what your saying here, could you show your steps explicitly?

3. May 18, 2008

### rosstheboss23

I assumed that velocity was the same for both around the wheels so since the v is the same for w1r1 and w2r2 where w2 is equal to 120rev/min which I convert to roughly 13 rad/s. The radii for both equations are given to be .11m for r1 and .25m for r2 so I solved for equation v2=w2r2 by plugging in 13(.25) and got roughly 3.25 m/s which I used for v1 and solved for angular speed in v1=w1r1. The angular speed for this I got to be roughly around 30rad/s. Then I used the kinematics equation V= Vo +at using 30rad/s for V and 0 for Vo and then for acceleration the 1.6 rad/s(squared) provided. Does this sound like a legitimate way of trying to solve the problem?

4. May 18, 2008

### Hootenanny

Staff Emeritus
Your method looks spot on too me

5. May 18, 2008

### rosstheboss23

Good. Thanks alot. I appreciate your help.

6. May 18, 2008

### Hootenanny

Staff Emeritus
A pleasure

Don't forget to mark the thread as solved when your done, thanks.