Linear Speed of Point on Pulley

KindofSlow

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Pulley (vertical disc): r = 0.33m, m = 4.00kg, initially at rest
Rope pulling (perpendicularly from side) with Force = F(t) = 3.00t - 0.20(t^2)
At t(in seconds) = 8, find linear speed of a point on the rim of the pulley.

Torque = R*F = 0.99t - 0.066(t^2) = 3.696 mN
From here, I don't know how to get either velocity or acceleration since rotational inertia is not given and acceleration is not constant.
I'm hoping someone can give me a little hint about what I'm missing and send me in the right direction.

Thank you
 
Pulley (vertical disc): r = 0.33m, m = 4.00kg, initially at rest
Rope pulling (perpendicularly from side) with Force = F(t) = 3.00t - 0.20(t^2)
At t(in seconds) = 8, find linear speed of a point on the rim of the pulley.

Torque = R*F = 0.99t - 0.066(t^2) = 3.696 mN
From here, I don't know how to get either velocity or acceleration since rotational inertia is not given and acceleration is not constant.
I'm hoping someone can give me a little hint about what I'm missing and send me in the right direction.

Thank you
You can compute the moment of inertia of a disk if you know m and R. Either find a formula (for instance in your text) or integrate.
 
Unfortunately, the disc is not uniform (apologies as I should have stated that up front) so I cannot use I = (1/2)m(R^2) (formula for disc or cylinder) and I cannot integrate as I do not have mass as a function of R.

Thank you
 
Unfortunately, the disc is not uniform (apologies as I should have stated that up front) so I cannot use I = (1/2)m(R^2) (formula for disc or cylinder) and I cannot integrate as I do not have mass as a function of R.
Is the pulley fastened down to a table or something, so that only rotational motion matters?

If you integrate the Torque from 0 to 8 s you will have a quantity with units on (N-m-s), which represents angular "impulse" or change of angular momentum. The units can be rewritten m^2-kg/s. You could divide by kg (e.g., mass=4 kg) and by a distance (e.g., radius = 0.33 m) resulting in units of linear velocity, but that would represent linear motion, not angular.

If you leave I as a parameter, you could get angular momentum as a function of I.

Is there any more you can tell us about the mechanical system? When you said perpendicular force I took that to be perpendicular to the axis of the pulley. Is that true?
 
I meant perpendicular to r (the lever arm) so that torque is just r*F without having to account for force at an angle to the lever arm.

Thank you
 
Regarding you first question: Is the pulley fastened down to a table or something, so that only rotational motion matters?

The axis of the pulley is horizontal and we are told to ignore friction (between axle and pulley).
So I think yes, only rotational motion matters.

Thank you
 
Dr. Phil,
I just discovered that we do know I from a previous problem with the same pulley, in which we were given rotational acceleration (so finding I was easy).
I didn't realize I was supposed to be carrying forward information from a previous problem - sorry about that.
Thanks for all your help.
 
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