Turntable Acceleration question

[Husker70]

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

 

[LowlyPion]

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..

 

[Husker70]

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

 

[LowlyPion]

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.

 

[Husker70]

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

 

[LowlyPion]

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.

 

[LowlyPion]

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 .

 

[Husker70]

Thanks,
That makes sense to me. Thanks for your time. I'm getting
ready to go take my first exam.

Take care,
Kevin

 

[LowlyPion]

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.