Finding radial & tangential acceleration at a point

  • #1

Homework Statement


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
(a) the radial acceleration,
(b) the tangential acceleration,
(c) the total acceleration of the point.

Homework Equations


a_r = v2/r
a_total = sqrt (a_t2 + a_rr)

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?
 

Answers and Replies

  • #2
264
26
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
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?
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement


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
(a) the radial acceleration,
(b) the tangential acceleration,
(c) the total acceleration of the point.

Homework Equations


a_r = v2/r
a_total = sqrt (a_t2 + a_rr)

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

12-rotational-motion-21-728.jpg


 
  • #5
264
26
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?
What does that imply about any changes might be taking place in the value of the radial acceleration?
 
  • #6
What does that imply about any changes might be taking place in the value of the radial acceleration?
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
20,291
4,290
The analysis in your original post was flawless. Nice job.
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?
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:
  • #8
264
26
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.
 

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