Angular speed of a disk, given a force

AI Thread Summary
A uniform-density 7 kg disk with a radius of 0.27 m is initially at rest and is subjected to a constant force of 45 N applied through a distance of 0.9 m via a string. The discussion revolves around calculating the angular speed of the disk using the work-energy theorem and relevant equations for rotational motion. Participants clarify the relationship between torque and angular acceleration, emphasizing the importance of correctly applying the equations for kinetic energy and moment of inertia. One user initially miscalculates by solving for the period instead of angular velocity but ultimately arrives at the correct answer after reevaluating their approach. The conversation highlights the significance of understanding the principles of rotational dynamics in solving such problems.
zalnas
Messages
9
Reaction score
0

Homework Statement


A uniform-density 7 kg disk of radius 0.27 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 45 N through a distance of 0.9 m. Now what is the angular speed?



Homework Equations


Krot = .5*I*omega2
Idisk = .5*m*R2
K = .5*m*v2
where I is the moment of intertia



The Attempt at a Solution



I've attempted setting Krot equal to the force applied times the distance over which it is applied. I then plugged in I using the given information, and solved for the period from the Krot equation.
 
Physics news on Phys.org
I think it will help you to realize a couple equations relating to torque

\tau = I \frac{d \omega}{dt}
\tau = \mathbf{r} \times \mathbf{F}

Though I am not entirely sure what the problem means by "and you pull on the string with a constant force of 45 N through a distance of 0.9 m" maybe you will and you will be on your way ;)
 
Thanks for the reply.

I tried setting the two equal:
.5*7*.272 d\omega/dt = .27*45

Which yielded 47.62 m/s. However, that isn't the correct answer.
 
Oh wait, my bad, I know what the line is alluding to now. Sorry for leading you astray. Yeah, you had the right idea before.

W=Fd
Krot=.5I\omega^2

Work energy theorem ought to work pretty well here. Seems you did that, so why did you solve for the period? Solve for angular velocity.
 
Ah, I'm not sure why I solved for the period. I got the correct answer, thank you very much!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top