Angular acceleration of a cylinder rolling up an inclined plane?

[khfrekek92]

Homework Statement

The cylinder is being pulled up the plane by a block. I've found that it IS rolling without slipping, and the acceleration is 5.00156 m/s^2.

Homework Equations

a=R[alpha]

The Attempt at a Solution

I've heard that the angular acceleration is the linear acceleration/radius, but this doesn't make sense, because using a radius of .2m, I get 25.0078 s^-2. What kind of units are inverse seconds squared? Does an angular acceleration of 25 even make sense for this situation? and is that even the right units? I thought it was supposed to be in rad/sec^2?

I've also seen that [torque]=I[alpha] Can I use that somehow? What would the torque be?

 

[khfrekek92]

Now I've tried using [torque]=rf and [torque]=I[alpha] and set them equal to each other and have gotten [alpha]=42.44.. How can this be right?? I'm getting two completely different answers :( :(

 

[Dick]

You haven't really described the whole problem so I don't know why you are getting two answers for the torque. But a radian angular measure is the distance along a circle divided by the radius of the circle. Since they are both measured in meters the units of radians are meters/meters. There's nothing wrong with writing rad/sec^2 but since 'rad' is formally dimensionless, they often omit writing 'rad'.

 

[khfrekek92]

Okay awesome! Well there are two inclined planes connected at the tips each with angle [theta] and [phi] which are 30 degrees, and 60 degrees respectively. On the 30 degree plane is a cylinder with mass 1.2 kg and radius .2m, and on the 60 degree plane is a block of mass 3 kg. These two objects are connected by a string running over a pulley. So the block is pulling the cylinder up its plane. The tension in the string is 16.97 N, and the acceleration (linear) is 5.00156 m/s^2. I'm not used to angular acceleration so I'm not sure if 20 rad/s^2 is acceptable for this situation... :/