# Solving for r: 3.24 Meters Too Far?

• Wzss

#### Wzss

Homework Statement
A bug crawls outward from the center of a CD spinning at 200 revolutions per minute. coefficient of static friction between bug's sticky feet and disc surface is 1.2. How far does the bug get from the center before slipping?
Relevant Equations
centripetal acceleration = (v^2 / r) = (4pi^2r)/ t^2
F_net= ma (newton's second law)
static friction = "mu" F_n (normal force)
200rpm/60 = 3.33 revolutions per second
I set (μ_s) *mg equal to (m*4pi^2r)/T^2
Then I solved for r
However, when I solve for r, I get 3.24 meters which seems much too far for a bug to travel. Should I have divided by something somewhere instead of multiplied?

I set (μ_s) *mg equal to (m*4pi^2r)/T^2
Then I solved for r
However, when I solve for r, I get 3.24 meters which seems much too far for a bug to travel. Should I have divided by something somewhere instead of multiplied?
Yes, that's too big.
Please post the details of the calculation.

• MatinSAR and Wzss
Yes, that's too big.
Please post the details of the calculation.
Thank you for the response!
Here is what I've been getting
r = (μ_s (mg)) * (t^2/ 4pi^2) (got this equation from setting equal static friction and centripetal force, then eliminating mass and solving equation for r)
r = (1.2*9.8*(3.3)^2) / 4pi^2
r = 3.24 meters
I know this makes no sense, but I'm not sure how else to arrange the equation

Thank you for the response!
Here is what I've been getting
r = (μ_s (mg)) * (t^2/ 4pi^2) (got this equation from setting equal static friction and centripetal force, then eliminating mass and solving equation for r)
r = (1.2*9.8*(3.3)^2) / 4pi^2
r = 3.24 meters
I know this makes no sense, but I'm not sure how else to arrange the equation
You should always verify the units of your proposed solution:$$r=(1.2) (9.8 m/s^2)(3.3/s)^2/(4\pi^2)=3.24 m/s^4$$which aren't the right units for a radius. Check your algebra!

• MatinSAR and Wzss
You should always verify the units of your proposed solution:$$r=(1.2) (9.8 m/s)(3.3/s)^2/(4\pi^2)=3.24 ms$$which aren't the right units for a radius. Check your algebra!
I'm confused. How did you get 3.24 ms?
When I do it, the m/s^2 from the 9.8 and the s^2 after squaring T cancel out, leaving meters. Since 9.8 is acceleration, shouldn't it be m/s^2?

I'm confused. How did you get 3.24 ms?
When I do it, the m/s^2 from the 9.8 and the s^2 after squaring T cancel out, leaving meters
I updated my post.

• Wzss
I updated my post.
Oh I see, thank you
but then, is my answer completely wrong? how do I get an answer that makes sense for radius?

Oh I see, thank you
but then, is my answer completely wrong? how do I get an answer that makes sense for radius?
As I said, check the algebra you used to find the radius. In your formula, should the square of the frequency ##3.3/s## be in the numerator or the denominator?

• MatinSAR and Wzss
As I said, check the algebra you used to find the radius. In your formula, should the square of the frequency ##3.3/s## be in the numerator or the denominator?
ah I see now, thank you so much!

• berkeman