Calculating Angular Acceleration with Friction on an Inclined Plane

  • Thread starter Thread starter ripper9100
  • Start date Start date
  • Tags Tags
    Rotation Slipping
AI Thread Summary
A solid uniform cylinder with a mass of 5 kg is being pulled up a 30-degree incline by a force of 45 Newtons. The discussion focuses on calculating its angular acceleration, with confusion around the moment of inertia and net torque. The moment of inertia can be either 0.5mr^2 or 0.5mr^2 + mr^2, depending on the chosen rotational axis, with the center of mass being recommended for clarity. The net torque is primarily influenced by friction, which acts to oppose the slipping between the cylinder and the incline. Understanding the direction of friction is crucial, as it opposes the potential motion of the surfaces in contact.
ripper9100
Messages
7
Reaction score
0

Homework Statement


A solid uniform cylinder with mass 5 kg is being pulled with a constant force of 45 Newtons up a 30 degree incline. The force is acting on the cylinders center and is parallel to the incline. What is its angular acceleration?

I have a good idea on how to do the problem by using Newtons second law for a rigid body. The only part I'm confused about is on what the moment of inertia should be. I think Its either (.5mr^2) or (.5mr^2 +mr^2)?

please help
 
Physics news on Phys.org
im also iffy on the net Torque. I am thinking that friction is the only force causing torque since the 45 N force is acting on the cylinders center.
am i right?
 
Yes. I agree.
 
thanks for the help
 
ripper9100 said:
I have a good idea on how to do the problem by using Newtons second law for a rigid body. The only part I'm confused about is on what the moment of inertia should be. I think Its either (.5mr^2) or (.5mr^2 +mr^2)?
The moment of inertia depends on what you are using as the rotational axis; either expression will work, if you're careful.

ripper9100 said:
im also iffy on the net Torque. I am thinking that friction is the only force causing torque since the 45 N force is acting on the cylinders center.
Again, it depends on the axis of rotation. If you are using the center of mass as the axis, then you are correct.

Even though you have a choice, I recommend using the center of mass of the cylinder as your axis--I think it gives the best understanding of what's going on.
 
what about the friction force is it up the incline or down the incline? I am confused about because usually the direction of friction force is opposite the direction of motion.
 
Remember that friction acts to oppose slipping between surfaces. If there were no friction, which way would the surfaces (cylinder bottom and incline) slip with respect to each other? Use that to figure which way friction must act.
 
Back
Top