Circular Motion of a turntable spinning counterclockwise

AI Thread Summary
The discussion revolves around calculating the time during the interval 0<t<T when the centripetal acceleration of a point on a turntable's rim equals half its initial value. The centripetal acceleration is defined as a=ω²r, and participants explore how the angular velocity (ω) must change to achieve this condition. It is determined that for ω² to be half its original value, ω must reduce to approximately 70.71% of its initial value, leading to the conclusion that this occurs at t=0.2929T. The conversation emphasizes understanding the relationship between angular velocity and centripetal acceleration in the context of rotational motion.
Dan22
Messages
11
Reaction score
0

Homework Statement


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



Homework Equations


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



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
 
Physics news on Phys.org
Dan22 said:

Homework Statement


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



Homework Equations


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



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

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?
 
PeterO said:
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?

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
 
Dan22 said:
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

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?
 
Dan22 said:
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

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

ω will be at half its original value at 1/2t. but what can i do with that?
 
Dan22 said:
wow, duh it will ω will be at half its original value at 1/2t. but what can i do with that?

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.
 
PeterO said:
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.

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.
 
i'm still having trouble figuring out what ω needs to be for ω^2 to be half the size
 
  • #10
Dan22 said:
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.

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
for ω^2 to be half its size would ω= [α/(2r)]^.5 ? or am i just thinking myself in circles?
 
  • #12
PeterO said:
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 ?

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
Dan22 said:
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?

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
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
Dan22 said:
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

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 = 0.707EDIT: 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
PeterO said:
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

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
Dan22 said:
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?

What is part b?

I am about to leave my computer for 36 hours, so won't reply immediately.
 
  • #18
PeterO said:
What is part b?

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

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,
 
Back
Top