What is the coefficient of static friction for a penny on a rotating wheel?

AI Thread Summary
The discussion focuses on calculating the coefficient of static friction for a penny on a rotating wheel. The penny slides off at a rotational speed of 0.825 revolutions per second, and the user attempts to derive the coefficient using centripetal acceleration formulas. The mistake identified is the failure to incorporate gravitational acceleration (g = 9.8 m/s²) into the calculation. The correct approach involves relating the centripetal force to the normal force to find the coefficient of static friction. The user acknowledges the oversight and seeks clarification on the correct method.
d.tran103
Messages
39
Reaction score
0
Can someone look at this and tell me what I'm doing wrong? Thanks!

Homework Statement


You place a penny on a level rotating wheel at a point 5.00 cm from the center of rotation. You increase the speed of rotation gradually, and once you reach a rotational speed of 0.825 revolutions each second, the penny slides off the wheel. What is the coefficient of static friction

a) 0.718

b) 0.409

c) 0.253

d) 0.137

Homework Equations


Fc = mv^2/r = mac
v=wr

The Attempt at a Solution


Fc = mv^2/r = mac

mv^2/r = mac turns to v^2/r = ac. I converted 0.825 rev/s to rad/s by multiplying by 2pi/1rev to get 5.18 rad/s. I plugged w into v=wr, v=(5.18 rad/s)(.05m) to get v=.259 m/s. I plugged v into v^2/r = ac, .259^2/.05 and got 1.34. However, 1.34 isn't an answer. Can someone tell me what I'm doing wrong?
 
Physics news on Phys.org
You found the acceleration. The question isn't asking for the acceleration. It's asking for a coefficient of static friction. That's the ratio between the normal force and the centripetal force. There should be a g=9.8m/sec^2 in there somewhere.
 
You solved for the centripetal acceleration. Now you need to find and identify the centripetal force.
 
Thanks, I was missing 1/9.8!
 
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