# Circular Motion of a turntable

1. Feb 21, 2013

### Dan22

1. The problem statement, all variables and given/known data
A turntable of radius "r" is spinning counterclockwise at an initial rate of ω. at t=0, its rotation rate begins to slow at a steady pace. the rotation finally stops at t=T. At what time during the time interval 0<t<T was the magnitude of the centripetal acceleration of a point on the turntable's rim equal to half its initial value? Express the answer in terms of T

2. Relevant equations
centripetal acceleration=ω^2r
ω=ω°+αt ?
Δθ=ω°t+.5αt^2

3. The attempt at a solution
got about a page and a half of hand written notes right here and nothing to show for it. not really sure where to start, and definitely not sure how to right equation in terms of T. any suggestions are appreciated

2. Feb 21, 2013

### PeterO

The expression for centripetal acceleration is where you need to start.

a = ω2r

you want that value to halve, but without the value of "r" changing, so what must happen to the value of ω and when will that happen?

3. Feb 21, 2013

### Dan22

i'm honestly not sure how to determine what ω needs to be nor when it will get there. ω^2 obviously has to be half as large but i'm not sure where to go from there

4. Feb 21, 2013

### PeterO

At what time will ω be 1/2 its original value?
At what time will ω be 1/3 its original value?
At what time will be 57% of its original value?

5. Feb 21, 2013

### PeterO

If ω2 is to be half as large, what must the value of ω be?

6. Feb 21, 2013

### Dan22

ω will be at half its original value at 1/2t. but what can i do with that?

7. Feb 21, 2013

### PeterO

1/2 is the easy one [though it should have been 1/2T I believe] Edit: and T/2 would be a better way of saying that.

try the other two.

8. Feb 21, 2013

### Dan22

do you mean when ω will be at 1/3ω and 57%ω. wouldn't that just be at T/3 and .57T. not sure why i would need those to be honest.

9. Feb 21, 2013

### Dan22

i'm still having trouble figuring out what ω needs to be for ω^2 to be half the size

10. Feb 21, 2013

### PeterO

I suspected you might say that. Both those answers are wrong.

Will the turntable have slowed to 1/3 its speed before or after it has slowed to half its speed?

is T/3 before or after T/2 ?

11. Feb 21, 2013

### Dan22

for ω^2 to be half its size would ω= [α/(2r)]^.5 ? or am i just thinking myself in circles?

12. Feb 21, 2013

### Dan22

man i'm feeling stupid right now, it would be at 1/3 of its speed at 2/3T and at 57% of its speed at .43T right?

13. Feb 21, 2013

### PeterO

That's better.

Now, suppose ω was originally 10. That would mean ω2 would be 100

What value would ω have to be for ω2 to be 50 ?

14. Feb 21, 2013

### Dan22

7.0711 instead of 10, and i didn't expect that to be a fixed relationship but it seems to be. so for ω^2 to be one half of that ω=.70711ω though i expect that there is a neater form of that

15. Feb 21, 2013

### PeterO

So W has to reduce to 70.71% of its original value. I am now confident you will be able to find when that happens.

By the way - the neater form is ω/√2 and of course 1/√2 = √2 / 2 and √2 / 2 = 1.414/2 = 0.707

EDIT: a quick course in "variation" - a mathematics topic - would be very handy at some stage. So much of Physics is dealt with in a Variation way.
eg: How does the speed of a satellite change of its radius of orbit is doubled. What is the value of g [9.8 at the surface of the Earth] if we move to a position 5R from the centre of the Earth. How about 5R from the surface of the Earth.

16. Feb 21, 2013

### Dan22

so the answer is t=.2929T is that correct. pretty sure it is, and thank you very much you were very helpful. willing to help with part b?

17. Feb 21, 2013

### PeterO

What is part b?

I am about to leave my computer for 36 hours, so won't reply immediately.

18. Feb 21, 2013

### Dan22

no problem, i think i got this it just has to do with finding at what time ω^2t is greatest pretty sure i got this, thanks for all your help,