# Finding radial & tangential acceleration at a point

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1. Nov 19, 2015

### vetgirl1990

1. The problem statement, all variables and given/known data
A point on a rotating turntable 20.0cm from the center accelerates from rest to final speed of 0.700m/s in 1.75s. At t=1.25s, find the magnitude and direction of
(b) the tangential acceleration,
(c) the total acceleration of the point.

2. Relevant equations
a_r = v2/r
a_total = sqrt (a_t2 + a_rr)

3. The attempt at a solution
I'm having trouble understanding non-uniform circular motion.

At t=1.25s, this is before the particle reaches it's final velocity at 1.75s. If I found average acceleration over that time period (a = 0.7m/s / 1.75s =0.4m/s/s), can I then use this to find the velocity at 1.25s, then find the radial acceleration from there?
a_avg = 0.4m/s/s
v = a*t = (0.4m/s/s)(1.25s) = 0.5m/s
a_radial = v2 / r = (0.52) / 0.2m = 2.5m/s/s at point 1.25s

I'm not sure if it's right to use this average acceleration applied to any point between when the particle goes from it's initial (v=0m/s) to final velocity (0.7m/s). Since the particle is accelerating, I know that of course, the velocity changes, but does the magnitude of acceleration (0.4m/s/s) change moving around the circle?

2. Nov 19, 2015

### J Hann

F = m a right?
What if the particle were held on the turntable by frictional force?
Suppose the frictional force is more than that required to provide the tangential acceleration.
What happens to the particle as the turntable continually speeds up?

3. Nov 19, 2015

### vetgirl1990

Well if the turntable continually speeds up, it may reach a speed that is fast enough that the frictional force is overcome and the particle accelerates off the turntable, tangent to the path?

4. Nov 19, 2015

### SteamKing

Staff Emeritus
Just like you have the SUVAT equations for linear motion, there is a similar set of circular motion equations:

5. Nov 19, 2015

### J Hann

What does that imply about any changes might be taking place in the value of the radial acceleration?

6. Nov 19, 2015

### vetgirl1990

Well I know that radial acceleration arises from the changes in the direction of the velocity vector... so I suppose the value of radial acceleration doesn't change at the instant the particle flies off the turntable? I'm not exactly sure...

7. Nov 19, 2015

### Staff: Mentor

The analysis in your original post was flawless. Nice job.
The problem statement implies that the tangential acceleration was constant over the 1.75 sec. See SteamKing's response in post #4.

Chet

Last edited: Nov 19, 2015
8. Nov 19, 2015

### J Hann

You had the right idea above when you mentioned "tangent to the path".
That's fine for an object on a string at moving "constant" speed" because there is no force in the tangential direction
and when the string breaks there are no forces acting on the string.
Here, you are asked to find two accelerations at right angles to each other, and then the resultant of these two accelerations.
I guess that your original question was can you use the average acceleration in the tangential direction.
The equations ("Steam King") above should answer that.
The question implies that the turntable is accelerating uniformly.