Physics Question Involving Angular Speed and Angular Velocity.

AI Thread Summary
The discussion revolves around calculating the angular speed and linear speed of a rotating slotted wheel used in a historical method for measuring the speed of light. The wheel has a radius of 5.6 cm and 75 slots, with a mirror located 800 m away. The initial calculations for angular speed yielded incorrect results due to premature rounding of time values. The correct approach emphasizes maintaining significant figures throughout the calculations to avoid errors. The conversation highlights the importance of precision in physics problems, particularly when dealing with constants like the speed of light.
McAfee
Messages
96
Reaction score
1

Homework Statement


An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through a slot at the outside edge of the wheel, as in Fig. 10-29, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.6 cm and 75 slots at its edge. Measurements taken when the mirror is L = 800 m from the wheel indicate a speed of light of 3.0 x 105 km/s. (a) What is the (constant) angular speed of the wheel? (b) What is the linear speed of a point on the edge of the wheel?

fizg0.gif



Homework Equations



t=(2L)/C

angular velcity= answer is part a *.056m

The Attempt at a Solution



For part (a)
I did use =angle/time

angle I solve (2Pi)/75 = .083776

I used t = 2L/c = 2(800)/(3x10^8) = .000005 sec

Then divided .083776/.000005 = 16755.2 rad/s

For part (b)

= my answer is part(a) * .056m

=938.291 rad/s

I know I'm doing something wrong, but don't know what it is. Can anyone help?
 
Physics news on Phys.org
I get 15,707 rad/s. You're introducing a huge error by prematurely rounding the value of the Δt to 5e-6 s when it should be 5.333333333333e-6 s. Keep all significant figures in intermediate calculation steps. Round only the final answer at the end.
 
You are right I rounded to early so it affected by final answer thank you very much and I hope you had a Happy Thanksgiving as well.
 
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