Does a point on the rim of the wheel have (a)tangential acceleration and (b) centripetal acceleration?

Question

A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant angular velocity. Does a point on the rim of wheel have (a) tangential acceleration and (b) centripetal acceleration? In each case, give your reasoning.

is this considered nonuniform circular motion because the tangential speed changes? So wouldnt there be both tangential and centrepetal accleration at all points on the rim?

Answer 1

The tangential speed DOESN'T change (it says in the problem 'constant angular velocity'), so it stays spinning at the same speed.

What's been done to analyze this problem is that one defines two coordinates, like x and y, with some major differences. The first is that one of these directions lies on the radius of a circle, and the other is perpendicular, tangent to it. Think about this for a second. That means that these axes are different for every angle! As something travels around in a circle, the coordinates travel with it. In the case of uniform circular motion, there is ALWAYS a force along R, and it's vectorally CONSTANT in these coordinates ONLY. Not so in x and y. Only if the thing is speeding up or slowing down in the angular (tangential) direction must we have a force along the tangent. In this case we don't.

Answer 2

No it does not have tangential acceleration. Yes its velocity vector is changing direction as it rotates. Remember acceleration is a change in velocity regardless of whether its due to a change in magnitude or direction. But the change in the velocity vector is due to the change in the direction of the centripetal acceleration vector not a tangential acceleration.

Yes any rotating point always has centripetal acceleration. If the magnitude of the centripetal acceleration vector is constant then the angular velocity of the point is also constant, If the magnitude of the centripetal acceleration vector is changing then the angular velocity will not be constant and the point will not be in uniform circular motion.

No, your example is still considered uniform circular motion. No there will not be tangential acceleration. Yes there will be centripetal acceleration at all points on the frame.

Bottom Line: The tangential velocity of a point in uniform circular motion always changes but its change is due to the centripetal acceleration not a tangential acceleration.

Answer 3

Typically the error comes in not understanding the terminology. Angular speed W = 2pi F in radians per second typically. F is the frequency of the rotation; in your case F = 40 rev/min, which must be converted to rev/second for consistency with the units you want in the answer. rev/min . min/60 sec ~ 1/60 rev/sec; so F = 40 rev/min = 40/60 rev/sec = 2/3 rev/sec is what you'd use. 1. The centripetal acceleration is Ap = V^2/R = W^2 R = (2pi F)^2 R = (2*pi*2/3)^2 * .2 = ? m/s^2 you can do the math. R = .2 m. Note that tangential speed V = WR, which is where W^2 R comes from. 2. More frequency stuff, as T = 1/F is the period. So when F = 2/3 rev/sec (also cycles/second as 1 revolution = 1 cycle) we have T = 1/F = 1/(2/3) = 3/2 seconds. ANS You would benefit from rereading your assignment and focusing on the definition of angular speed W and how frequency (rev/sec or cps) relates to it.