# Homework Help: Acceleration and Rotational Motion

1. Oct 11, 2007

### Eternalmetal

1. The problem statement, all variables and given/known data
If a rotational object is moving from rest to 200rpm in .23 seconds, the acceleration is?

2. Relevant equations
a = V^2/r
v = 2Pi(r)/t
I am sure there are others, but this is just a General Physics I course, so its dealing with Newton and Kepler's laws.

3. The attempt at a solution

To be honest, I dont know if it is totally solvable. I have a feeling that I am going to be left with a variable in my answer since the radius is not given. Without a radius, I cant figure out velocity, and without velocity I cant figure out acceleration. Judging by previous questions of my professor, the answer would most likely be in meters/second.

Working with radius as a variable, I am getting something like this:
a = (2π /.23)² r

Would this be logical, or am I going about this in the wrong way?

Last edited: Oct 11, 2007
2. Oct 11, 2007

### Staff: Mentor

This problem is asking for angular acceleration, which is measured in radians/s^2. Angular acceleration and velocity are completely analogous to linear acceleration and velocity--the same kinematic equations apply. What's the definition of acceleration? (Be sure to convert from rpm to rad/sec.)

3. Oct 11, 2007

### Eternalmetal

Ok, so 200 RPM divided by 60 is 3.3 radians/sec (correct?). Since I have the units of radians per second, I should just multiply by 1/t to get an answer in radians/sec², correct?

3.3 radians/sec * 1/.23s = 14.35 radians/sec²

Does this sound a little better then?

4. Oct 11, 2007

### Staff: Mentor

No. Rpm stands for revolutions per minute, not radians per minute. How many radians are in one complete revolution?

But once you get the correct value for rad/s, you have the right idea for finding the average acceleration.

5. Oct 11, 2007

### Eternalmetal

Ah, if im not mistaken, would it be 2Pi radians in one revolution? So I would multiply 3.3 revolutions/sec by 2Pi to get about 20.73 radians/sec, and then divide by .23 like I did previously and get a final value of 90.15 radians/sec² ?

Thanks for your help Doc Al.

6. Oct 11, 2007

### Staff: Mentor

Exactly right. (But carry out your intermediate calculations with a bit more accuracy. 3.3 should really be 3.3333... )

7. Oct 11, 2007

### Eternalmetal

My mistake, I guess I was being a little careless. Thanks again for your help, we definitely did not go over this in lecture. Now that I realize what this is, it is a chapter that we did not cover yet, we were actually going over circular motion instead.

Last edited: Oct 11, 2007