Circular motion ultra-centrifuge spin

AI Thread Summary
An ultra-centrifuge with a cylindrical disk is analyzed for its deceleration after the driving force is turned off, with an initial angular speed of 4.50 x 10^5 rad/s and a deceleration of 0.390 rad/s². The discussion focuses on calculating the time until the rotor comes to rest, the number of revolutions during this period, and the initial linear, radial, and tangential accelerations at the disk's edge. Participants suggest using angular measures and encourage deriving equations from angular motion graphs for clarity. Overall, the conversation emphasizes the importance of understanding constant deceleration in circular motion. The thread seeks to clarify the application of physics equations in this context.
Jabababa
Messages
52
Reaction score
0

Homework Statement


An ultra-centrifuge has a cylindrical disk mounted on an axle that is almost frictionless. The disk spins about an axis through its centre as shown. If the disk is spinning with an angular speed of 4.50 x 10^5 rad/s and the driving force is turned off, its spinning slows down (due to air resistance) at a rate of 0.390 rad/s^2.

a) How long does the rotor spin before coming to rest?

b) during the time that it is slowing down, how many revolutions does the rotor spin before coming to rest?

c) If the disk has a diameter of 26.0 cm, find the initial linear speed of a point on the outer edge of the disk.

d) find the magnitude of the initial radial acceleration of a point on the outer edge of the disk when it first starts to slow down.

e) find the magnitude of the initial tangential acceleration of a point on the outer edge of the disk when it first starts to slow down.


Homework Equations





The Attempt at a Solution



a) Vf= Vi + at => t = Vf-Vi/a

is that correct? I will start off with 1st one, and will add the work i did on b-e. If someone can let me know if that is correct. If wrong please explain what i did wrong.
 
Physics news on Phys.org
Jabababa said:

The Attempt at a Solution



a) Vf= Vi + at => t = Vf-Vi/a

is that correct? I will start off with 1st one, and will add the work i did on b-e. If someone can let me know if that is correct. If wrong please explain what i did wrong.

You will be better off working with the angular measures for this problem. The same formula can be applied to the angular measures by recognizing the analogs: d → θ ; v → ω ; a → ##\alpha##.
 
Other than the variables, am i on the right track?
 
try it and see ... starting calculation without being sure you are on the right track is good practice.

you can check also your reasoning ... the situation involves constant deceleration - that sort of motion has a special name.

if you are uncertain about equations, try deriving them from a ##\small \omega## vs t graph.
 
Jabababa said:
Other than the variables, am i on the right track?

You are, but it's a little early in the game to tell where the track might lead :smile:

Show some more work and a result.
 
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