Circular Motion: Calc Centripetal Force on Amoeba

In summary, the centrifuge pushes an amoeba down at a constant velocity. The force on the amoeba is due to the centripetal force and is 9.87x10^5Newton.
  • #1
brentwoodbc
62
0

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.
 
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  • #2
brentwoodbc said:
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
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
brentwoodbc said:

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
Doc Al said:
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
brentwoodbc said:
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
Doc Al said:
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.
 

1. What is circular motion?

Circular motion is a type of motion in which an object follows a circular path around a fixed point. It can also be described as the motion of an object at a constant speed along a circular trajectory.

2. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directed towards the center of the circle. It is responsible for keeping the object in its circular motion and preventing it from flying off in a straight line.

3. How is centripetal force calculated?

The formula for calculating centripetal force is F = m x v^2 / r, where F is the centripetal force, m is the mass of the object, v is the velocity of the object, and r is the radius of the circular path.

4. Why is centripetal force important in circular motion?

Centripetal force is important in circular motion because it is the force that allows objects to move in a circular path without deviating from it. Without centripetal force, objects would move in a straight line, and circular motion would not be possible.

5. How is centripetal force related to amoebas?

In the context of amoebas, centripetal force is important because it is responsible for keeping the amoeba's cell membrane intact as it moves and changes shape. This allows the amoeba to move and feed efficiently, as well as maintain its structural integrity.

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