# Angular acceleration and newton's 2nd law

1. Jun 1, 2007

### Bjamin0325

1. The problem statement, all variables and given/known data
A solid cylindrical disk has a radius of 0.16 m. It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a 45 N force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of 110 rad/s2. What is the mass of the disk?

F=45 N
r=0.16 meters
a=110
m=?

2. Relevant equations

torque=F(l)
torque = force * lever arm

net external torque=mr^2*a
3. The attempt at a solution
So far, I've gotten torque = 45*.16 = 7.2
7.2=m(0.16)^2 * 110
and solving for m gives me 2.556
but that is not correct. I am absolutely stumped.

Thanks for any help.

2. Jun 1, 2007

### Staff: Mentor

That's not true in general, and certainly not true for a cylinder. Look up the rotational inertia of standard shapes.

3. Jun 1, 2007

### D H

Staff Emeritus
What makes you think this? What is the moment of inertia of a solid cylinder about its axis of rotation?

Dang Doc, you type too fast.

4. Jun 1, 2007

### Bjamin0325

My book doesn't have anything about different moments of inertia for different shapes, so I figured that was right.

I looked up the rotational inertias :http://en.wikipedia.org/wiki/List_of_moments_of_inertia

and it says for a thin solid disk, it is I=(mr^2)/4.

I tried this:
net torque=I*a, so
7.2=((mr^2)/4)*110, and solving for m did not give me the right answer.

oh no.

5. Jun 1, 2007

### Bjamin0325

nevermind, I=mr^2/2.
thanks for the help, everybody.