Grindstone angular acceleration

AI Thread Summary
The discussion revolves around calculating the angular acceleration of a disk-shaped grindstone after power is shut off. The grindstone has a mass of 1.69 kg and a radius of 7.95 cm, initially rotating at 707 revolutions per minute. It takes 32.1 seconds to come to a stop, indicating a need for constant angular acceleration calculation. The initial attempt at solving the problem involved incorrect assumptions about energy and rotational motion equations. The correct approach emphasizes converting the initial angular velocity from revolutions per minute to radians per second and using the appropriate kinematic equation to find the angular acceleration.
Vanessa23
Messages
41
Reaction score
0
[SOLVED] Grindstone angular acceleration

Homework Statement


Object: a disk-shaped grindstone of mass 1.69 kg and radius 7.95 cm that operates at 707 rev/min. When the power is shut off, you time the grindstone and find it takes 32.1 s for it to stop rotating. What is the angular acceleration of the grindstone? (Assume constant angular acceleration.)

Homework Equations


W=.5mv^2
V=distance/t
W=Wo+at
a=alpha, which is angular acceleration

The Attempt at a Solution


V=[707*(2pi/60)]/32.1 = 2.31m/s
W=0
Wo=.5(1.69)(4.50^2) = 4.50J
0=4.50+a(32.1)
therefore, a=-.140 rad/s^2

however, this is the wrong answer according to the program. It didn't have a problem with the units though, so I don't know what I am doing wrong. Please point out any wrong assumptions or completely wrong or silly mistakes. THANK YOU!
 
Physics news on Phys.org
You seem to have W representing two different things here. I don't know what you are trying to do with the first equation you have, but it's not for rotational motion. Also, Wo is your initial angular velocity, I don't understand why you are trying to make it into energy?:confused:

All you need for this is your third equation. Change the 707 rev/min into rad/s, that is your Wo. Solve for the angular acceleration.
 
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