Find its angular acceleration in radians per second per second

AI Thread Summary
To find the angular acceleration of a potter's wheel that moves uniformly from rest to an angular speed of 0.23 revolutions per second in 29 seconds, one must first define angular acceleration as the change in angular velocity over time. The conversion from revolutions per second to radians per second involves multiplying by 2π, resulting in an angular speed of approximately 1.44 radians per second. The angular acceleration can then be calculated using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. Participants are encouraged to show their work and follow a posting template for clarity. Understanding these concepts is essential for solving the problem effectively.
cake90036
Messages
2
Reaction score
0
A potter's wheel moves uniformly from rest to an angular speed of 0.23 rev/s in 29 s.
 
Physics news on Phys.org
Anybody
 
cake90036 said:
A potter's wheel moves uniformly from rest to an angular speed of 0.23 rev/s in 29 s.
Welcome to PF.

You have to show us what you have done first. You should follow the posting template.

What is the definition of angular acceleration? How do you convert revolutions per second to angular speed?

AM
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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