Radial and tangential acceleration?

Question

A point on rotating turntable 21.5 cm from the center accelerates from rest to a final speed of .640 m/s in 1.70s. At t=1.28 s, find the magnitude and direction of each of the following:
(a) the radial acceleration
(b) the tangential acceleration

I'm not sure how to figure out the radial acceleration since this in not uniform circular motion. Thanks!

Answer 1

At t = 1.28 sec, the turntable is still accelerating; so a = alpha*R = v/t; where R = 21.5 cm from the center and alpha is the angular acceleration. When t = 1.28, we have v = alpha*R*t; so we need to find alpha. alpha = V/TR where V = .64 m/sec and T = 1.70 sec.

a) Therefore v = (V/TR)Rt = V(t/T) which is the tangential velocity at time t = 1.28 sec. Radial acceleration is a(t) = v^2/R = V^2(t/T)^2/R; where V = .64 m/sec, t = 1.28 sec, T = 1.7 sec, and R = 21.5 cm = .215 m. You can do the math.

b) Tangential acceleration is just v/t = V(t/T)/t = V/T, which is what it was to reach full speed V. Again, you can do the math.

Answer 2

"Constant angular velocity" means no angular acceleration, since acceleration is CHANGE in velocity with time. However, there IS tangential acceleration, because the velocity includes direction, and the direction component of tangential velocity is changing. Since the change in direction of tangential velocity is in the radial direction, there is also radial acceleration. If the angular velocity is increasing, then there is angular acceleration, as well as tangential and radial acceleration. The magnitude of linear acceleration would change if there is angular acceleration. Linear acceleration is proportional to angular velocity, which will be changing if there is angular acceleration.

Answer 3

It does not matter that circular motion is non-uniform. The usual formula
a_r = v²/R = ω²R = ωv
for radial accleration is still applicable.

Mainly because
a_r(t) = ω(t) x v(t)