• Support PF! Buy your school textbooks, materials and every day products Here!

Circular motion?

  • #1

Homework Statement



a test tube rotates in a centrifuge with a period of 1.2x10^-3s. The bottom of the test tube travels in a circular path of radius .15 m. with the centripetal force on a 2.00x10^-8kg amoeba at the bottom of the tube.




The Attempt at a Solution



ac=v^2/r=(4pi^2r)/T
cross multiplied and got.
v^2x(1.2x10^-3)=4pi^2x(1.5^2)
divided
and solved for velocity and I got.
V=27.21



then fc=(m2v^2)/r
fc=(2x10^-8)x(27.210^2)/.15
fc=9.87x10^5????
supposed to be 8.22x10^2


THanks.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
ac=v^2/r=(4pi^2r)/T
cross multiplied and got.
v^2x(1.2x10^-3)=4pi^2x(1.5^2)
divided
and solved for velocity and I got.
V=27.21
Two problems:
(1) That equation is not quite right. The right hand side should be: (4pi^2r)/(T^2)
(2) Why did you solve for V? What you want is v^2/r, which is given directly by the (corrected) right hand side.
 
  • #3
137
4
It's not clear what the question is asking.

If you're looking for the force then you could start by finding the linear velocity. Just think distance divided by time and the circumference of the circle that the end of the tube is moving along.

You have a formula for the acceleration in terms of v and r. Compare this to Newton's second law and you should be able to get the accelerating (centripetal) force in terms of v and r as well.
 
  • #4
LowlyPion
Homework Helper
3,090
4

Homework Statement



a test tube rotates in a centrifuge with a period of 1.2x10^-3s. The bottom of the test tube travels in a circular path of radius .15 m. with the centripetal force on a 2.00x10^-8kg amoeba at the bottom of the tube.

The Attempt at a Solution



ac=v^2/r=(4pi^2r)/T
cross multiplied and got.
v^2x(1.2x10^-3)=4pi^2x(1.5^2)
divided
and solved for velocity and I got.
V=27.21

then fc=(m2v^2)/r
fc=(2x10^-8)x(27.210^2)/.15
fc=9.87x10^5????
supposed to be 8.22x10^2

THanks.
Or more directly for the same result

F = m*ω2*r

where ω = 2π/T

F = m*(2π/T)2*r

Edit: I think the correct answer should have a (-) exponent ?
 
Last edited:
  • #5
Two problems:
(1) That equation is not quite right. The right hand side should be: (4pi^2r)/(T^2)
(2) Why did you solve for V? What you want is v^2/r, which is given directly by the (corrected) right hand side.
Thanks
You are correct, The T^2 was my mistake. I have the right answer now (8.22 x 10^-2)

I do not understand what you mean by just using the right side?
ac=(4pi^2r)/T
there's no v there.
I solved for v to use the formula fc = (m2v^2)/r
 
  • #6
Doc Al
Mentor
44,882
1,129
Thanks
You are correct, The T^2 was my mistake. I have the right answer now (8.22 x 10^-2)
I didn't check your calculation. Was it just a typo?

I do not understand what you mean by just using the right side?
ac=(4pi^2r)/T
there's no v there.
I solved for v to use the formula fc = (m2v^2)/r
You started with the equation: ac=v^2/r=(4pi^2r)/T^2
What you need (to move to the next step) is v^2/r, which equals (4pi^2r)/T^2. You don't need to know V explicitly:
ac=v^2/r=(4pi^2r)/T^2

thus:
Fc = mac = mv^2/r= m(4pi^2r)/T^2

You could go right to the answer using only r and T, which were given.
 
  • #7
I didn't check your calculation. Was it just a typo?


You started with the equation: ac=v^2/r=(4pi^2r)/T^2
What you need (to move to the next step) is v^2/r, which equals (4pi^2r)/T^2. You don't need to know V explicitly:
ac=v^2/r=(4pi^2r)/T^2

thus:
Fc = mac = mv^2/r= m(4pi^2r)/T^2

You could go right to the answer using only r and T, which were given.
Oh in the sense f=ma. that makes sense.
Thank you.
 

Related Threads for: Circular motion?

  • Last Post
Replies
8
Views
656
  • Last Post
Replies
4
Views
577
Replies
3
Views
4K
  • Last Post
Replies
18
Views
569
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
815
  • Last Post
Replies
7
Views
751
  • Last Post
Replies
1
Views
599
Top