Angular Acceleration Homework: 1014.65°, 17.97 rad/s

AI Thread Summary
The discussion revolves around a physics homework problem involving a wheel rotating with constant angular acceleration. The calculations show that the wheel rotates an angle of 1014.65 degrees between t=0 and t=3.67 seconds, and its angular speed at 4.1 seconds is 17.97 rad/s. Some participants question the correctness of the initial calculations, specifically regarding the need to square the time variable in the angular displacement formula. There is also confusion about tangential and total acceleration values, suggesting a possible mix-up with a different problem. The conversation highlights the importance of correctly applying formulas in angular motion problems.
dr2112
Messages
9
Reaction score
0

Homework Statement


wheel rotates w constant acceleration 3.65rad/s2
angular speed is 3.00 rad/s at t=0
a)what angle does wheel rotate between t=0 and t=3.67s?
b)what is angular speed of wheel at 4.1s?


Homework Equations


wt+1/2alphat
(3.00rad/s)(3.67t)+1/2(3.65rad/s2)(3.67)2=17.70rad
17.70rad*360degrees/628rad=1014.65degrees

w=w0+alpha t
3.00rad/s+3.65rad/s2(4.1s)=17.97rad/s

The Attempt at a Solution


a=1014.65 degrees
b=17.97rad/s

 
Physics news on Phys.org
Hi dr2112,

dr2112 said:

Homework Statement


wheel rotates w constant acceleration 3.65rad/s2
angular speed is 3.00 rad/s at t=0
a)what angle does wheel rotate between t=0 and t=3.67s?
b)what is angular speed of wheel at 4.1s?


Homework Equations


wt+1/2alphat
(3.00rad/s)(3.67t)+1/2(3.65rad/s2)(3.67)2=17.70rad

I don't believe this is correct. The final t needs to be squared:

<br /> \Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2<br />
 
alphysicist said:
Hi dr2112,



I don't believe this is correct. The final t needs to be squared:

<br /> \Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2<br />

Ok how is 1.60 m/s^2 for tangential and 8.61 m/s^2 for total acceleration?
 
dr2112 said:
Ok how is 1.60 m/s^2 for tangential and 8.61 m/s^2 for total acceleration?

Is this for a different problem, or is there more to the problem in your first post?
 
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