Static Friction and Uniform Circular Motion

AI Thread Summary
The discussion revolves around calculating the least coefficient of static friction required to keep a cat on a merry-go-round. The cat is positioned 4.0 meters from the center and completes a rotation every 6.3 seconds, leading to a calculated velocity of 3.99 m/s and a centripetal acceleration of 3.98 m/s². Participants clarify that the static friction force provides the necessary centripetal force, and the normal force equals the cat's weight (mg). It is emphasized that the mass of the cat is not needed for the calculation since the forces can be equated directly. The thread concludes with the realization that the centripetal acceleration was misinterpreted as a force, highlighting the importance of understanding the relationship between these concepts.
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Homework Statement



A cat sits on a merry go round at a radius of 4.0m from the center. The ride makes one complete rotation every 6.3 seconds. What is the least coefficient of static friction to keep the cat in place?

Homework Equations


Circumference = 2Pi*r
V=d/t
a=v^2/r


The Attempt at a Solution


I first figured out the circumference of the circle traveled by the cat to be:

2 x 4.0 x Pi= 25.13m

Velocity of the cat:

V = 25.13/6.3 = 3.99 m/s

Centripetal acceleration of the cat:

a = 3.99^2 / 4.0 = 3.98 m/s^2

I'm assuming that i have to use the formula for static friction in some way, but i am not sure how to apply it to what i have calculated so far. Any help would be greatly appreciated, thanks.
 
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harpua09 said:

The Attempt at a Solution


I first figured out the circumference of the circle traveled by the cat to be:

2 x 4.0 x Pi= 25.13m

Velocity of the cat:

V = 25.13/6.3 = 3.99 m/s

Centripetal acceleration of the cat:

a = 3.99^2 / 4.0 = 3.98 m/s^2

Your answer here is correct, but you could have alternatively just found ω since they said "1 revolution in 6.3s" and then use a=ω2r


harpua09 said:
I'm assuming that i have to use the formula for static friction in some way, but i am not sure how to apply it to what i have calculated so far. Any help would be greatly appreciated, thanks.

Well the force of static friction is providing the centripetal force needed to keep the cat in circular motion. If the mass of the cat is 'm', then its weight is 'mg' and the normal reaction is 'N'. The frictional force is therefore F= μN.

Just equate the two and solve.
 
I'm not sure what you are referring to with "ω". I am also not sure hot to find N, if i do not know the mass of the cat.
 
Interesting. Seems like mass is indeed needed.
 
harpua09 said:
I'm not sure what you are referring to with "ω". I am also not sure hot to find N, if i do not know the mass of the cat.

ω=angular velocity, you did not need to use it, but it would have just been easier instead of finding the velocity and then using v2/r

fawk3s said:
Interesting. Seems like mass is indeed needed.

Not really. Remember, the cat is not moving upwards so the normal reaction N is just the weight 'mg'. N=mg.

and if Fcentripetal=Ffriction mass is not needed :wink:
 
Figured it out, thanks!
 
rock.freak667 said:
Not really. Remember, the cat is not moving upwards so the normal reaction N is just the weight 'mg'. N=mg.

and if Fcentripetal=Ffriction mass is not needed :wink:

Oh, my bad. Sorry. I seem to be missing a lot lately. For some fudgy reason my head thought of the centripetal acceleration as the force (without multiplying with mass).

/facepalm
 

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