Angular acceleration of a cylinder?

[khfrekek92]

Homework Statement

After some long calculations, I found that the linear acceleration of a cylinder on an inclined plane (attached to a block over a pulley on another inclined plane) was 8.76238 m/s^2. How do I find the angular acceleration of this rolling cylinder?

Homework Equations

I[omega]=mR^2[omega] maybe?

The Attempt at a Solution

I literally have no idea how to finish this problem.. Thanks in advance for any help!

 

[eczeno]

if the cylinder is rolling without slipping, then the angular acceleration is equal to the linear acceleration divided by the radius of the cylinder.

cheers

 

[Fewmet]

Homework Equations

I[omega]=mR^2[omega] maybe?

Isn't the moment of inertia for a cylinder rotating around ax axis through its length 1/2 mr² (or am I misremembering?). If so the equation you suggest cannot be correct.

You might make use of
a= r \alpha
where a is the tangential acceleration of a point on the circumference. Do you have the information to find that? Perhaps from the distance and time, if there is no slipping?

How about \tau=I\alpha and \tau=rF?

 

[cupid.callin]

Isn't the moment of inertia for a cylinder rotating around ax axis through its length 1/2 mr² (or am I misremembering?). If so the equation you suggest cannot be correct.

You might make use of
a= r \alpha
where a is the tangential acceleration of a point on the circumference. Do you have the information to find that? Perhaps from the distance and time, if there is no slipping?

How about \tau=I\alpha and \tau=rF?

Are you planning to find torque then angelar acceleration? ... that is surely not a good method
a = \alpha r is perfectly correct

its just a derived eqn from s = \theta r ... here s is linear displacement ...

 

[khfrekek92]

Okay so if [alpha]=a/r, I plug everything in and with r=.2m and a=5.00156 m/s^2, I get [alpha]=25.0078 s^-2. Inverse squared seconds? Is that a unit of angular acceleration? How would I get that into rad/sec^2? Just multiply it by 2[pi]? I think these angular accelerations are too high for a 5m/s^2 acceleration?

 

[cupid.callin]

no need to multiply be 2 pi
rad is not a dimensional unit
it is for pure numbers

angular acceleration is 25 rad/s^-2

and for your question that this looks too large for 5m/s^2 acceleration ...

consider 2 cylinders ... one of very smaller radius than other
in order to move together ... smaller one has to spin much more quickly
so even though their linear acceleration may be same ... due to small size ... smaller one will have much larger angular acceleration

 

[khfrekek92]

Oh okay awesome! Thank you so so much!