Turntable Acceleration question

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Homework Help Overview

The problem involves a point on a rotating turntable that accelerates from rest to a specific final speed over a set time period. Participants are tasked with finding the radial, tangential, and total accelerations at a given moment during this acceleration process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of speed at a specific time and the method for determining radial and tangential accelerations. Questions arise about the assumptions of constant acceleration and the relationships between different types of acceleration.

Discussion Status

Some participants have provided guidance on calculating tangential acceleration and have confirmed the radial acceleration found by others. There is ongoing exploration of the relationships between the different accelerations, but no consensus has been reached on all aspects of the calculations.

Contextual Notes

Participants are working under the assumption of uniform acceleration, and there is some uncertainty regarding the interpretation of the problem setup and the calculations involved.

Husker70
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Homework Statement


A point on a rotating turntable 20.0 cm from the center accelerates from rest to a final
speed of .700m/s in 1.75s. At t=1.25s, find the magnitude and direction of (a) the radial
acceleration, (b) the tangential acceleration, and (c) the total acceleration at that point


Homework Equations





The Attempt at a Solution


I get the speed by comparing the the speed at 1.75s to speed at 1.25. I get
this to be .5m/s
To find the radial acceleration at that point I do (.5m/s)2/.20m = 1.25m/s2
Is the tangential acceleration at that point 1.25/1.25? Which is the radial/time?
Kevin
 
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Husker70 said:

Homework Statement


A point on a rotating turntable 20.0 cm from the center accelerates from rest to a final
speed of .700m/s in 1.75s. At t=1.25s, find the magnitude and direction of (a) the radial
acceleration, (b) the tangential acceleration, and (c) the total acceleration at that point

Homework Equations



The Attempt at a Solution


I get the speed by comparing the the speed at 1.75s to speed at 1.25. I get
this to be .5m/s
To find the radial acceleration at that point I do (.5m/s)2/.20m = 1.25m/s2
Is the tangential acceleration at that point 1.25/1.25? Which is the radial/time?
Kevin

The tangential acceleration at that point is the change in tangential velocity. That would be the same acceleration you used to determine that it was moving at .5m/s namely (.7m/s) /1.75s

Presumably you know to sum the accelerations as vectors for the final answer..
 
Thanks but that what I'm unsure of. I simply assumed that this is a constant acceleration
and so by simple comparing taking 1.75/.7=1.25/x, I get .5 m/s
Did I start that wrong?

Kevin
 
Husker70 said:
Thanks but that what I'm unsure of. I simply assumed that this is a constant acceleration
and so by simple comparing taking 1.75/.7=1.25/x, I get .5 m/s
Did I start that wrong?

Kevin

You should expect that the acceleration is uniform, and that looked fine.

They should have said if it were not.

Good Luck.
 
Is that tang acceleration correct by using the radial acceleration at that point/time?
So that 1.25m/s2/ 1.25s = 1 m/s2 for tang?

Thanks,
Kevin
 
Husker70 said:
Is that tang acceleration correct by using the radial acceleration at that point/time?
So that 1.25m/s2/ 1.25s = 1 m/s2 for tang?

Thanks,
Kevin

Oops the acceleration is inverted. It's .7/1.75. Sorry for swapping the two.
 
Last edited:
That means you have a forward acceleration of .7/1.75 = .4m/s2

And your radial acceleration is 1.25m/s2 as you already found.

The angle then is tan-1(.4/1.25) = 17.74 degrees (ratio of Opposite/adjacent with respect to forward of the radius.)

The magnitude is the RSS of .4 and 1.25 .
 
Thanks,
That makes sense to me. Thanks for your time. I'm getting
ready to go take my first exam.

Take care,
Kevin
 
Husker70 said:
Thanks,
That makes sense to me. Thanks for your time. I'm getting
ready to go take my first exam.

Take care,
Kevin

Best of luck then.
 

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