How to find angular acceleration given a force applied?

[xtreameprogram]

1. The problem statement, all variables, and given/known data
Joe is painting the floor of his basement using a paint roller. A roller has a mass of 2.4kg and a radius of 3.8cm. In rolling the roller across the floor, Joe applies a force F= 16N at an angle of 35 degrees. What is the magnitude if the angular acceleration of the roller?

Homework Equations

I think F=ma and a_tan = r * a_rot

The Attempt at a Solution

I tried using F=ma to figure out the acceleration of the whole object which would then be equal to the tangential acceleration. I just then divided that by r to get a_rot

Fsinθ/2.4kg = a_tan = 3.824 m/s²

a_tan / r = 10.06 rad/s²

I am really not sure if this is correct or not. If this isn't then I feel like I would have to use torque, but I don't know the moment of inertia for it to solve for a_rot. Any help is appreciated, thank you.

 

[Arman777]

You can think paint roller as a cylinder.And you can calculate its moment of inertia (Or you search).After that using relationship between τ,I and ∝(angular acceleration).you can find ∝.

 

[haruspex]

Further to Arman's reply:
There are two ways to analyse this.
1. Take the centre of the roller as the axis. In this approach you use the moment of inertia about that axis, and must include the friction from the floor in the torque equation. So you get a torque equation for rotational acceleration and a force equation for linear acceleration. Each includes the friction.
2. Take the point of contact with the ground as the axis. This allows you to ignore friction for the rotational acceleration because the friction has no torque about that axis. In this approach you must use the moment of inertia of the cylinder about that point of contact. Use the parallel axis theorem.

 

[Arman777]

1. Take the centre of the roller as the axis. In this approach you use the moment of inertia about that axis, and must include the friction from the floor in the torque equation. So you get a torque equation for rotational acceleration and a force equation for linear acceleration. Each includes the friction.

Theres no given $μ_k$ Is that means we have to use your second approach...Or does the question says ignore friction ? (unlikely)

 

[haruspex]

Theres no given $μ_k$ Is that means we have to use your second approach...Or does the question says ignore friction ? (unlikely)

It doesn't say, but you have to assume it is rolling contact. (That means it would be $μ_s$, but even then that would only tell you the maximum frictional force. )
As I wrote, with method 1 you would have two equations involving frictional force, so you can eliminate it and or determine it. The coefficient is irrelevant.

 

[xtreameprogram]

Further to Arman's reply:
There are two ways to analyse this.
1. Take the centre of the roller as the axis. In this approach you use the moment of inertia about that axis, and must include the friction from the floor in the torque equation. So you get a torque equation for rotational acceleration and a force equation for linear acceleration. Each includes the friction.
2. Take the point of contact with the ground as the axis. This allows you to ignore friction for the rotational acceleration because the friction has no torque about that axis. In this approach you must use the moment of inertia of the cylinder about that point of contact. Use the parallel axis theorem.

I'll have to use the first method considering we haven't touched the parallel axis theorem. It isn't even in my textbook. Thank you