Heres a tough one i had on my final im dying to find out how to do it

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The discussion centers on calculating the centripetal acceleration of a mass rotating on a horizontal surface with air friction. The mass is 3 kg, the radius is 2 m, and the initial velocity is 10 m/s. The air friction is modeled by the equation 0.2v, leading to a first-order differential equation of the form -kv = mdv/dt. To find the centripetal acceleration after 5 seconds, one must solve this differential equation while applying the appropriate boundary conditions.

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I don't remember the question verbatim, but here is all the info

Homework Statement



There is a mass attached to a rope which is attached to a pole. It is set to rotate on a horizontal surface (basic circular motion scenario). There is no kinetic friction along the table, but there is air friction.

The magnitude of air friction is related to velocity with the equation .2v.
m=3 kg
r=2 m
v(initial)=10 m/s

What is the centripetal acceleration after 5 s.

Homework Equations



F=ma=m(v^2/r)

The Attempt at a Solution



i know you got to integrate the air friction function over the 5 s interval, but i can't express it in terms of t.

I am completely lost, please help. I know I got it wrong, but i really want to know how to do it.

Thanks!
 
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xxacefirexx said:
I don't remember the question verbatim, but here is all the info

Homework Statement



There is a mass attached to a rope which is attached to a pole. It is set to rotate on a horizontal surface (basic circular motion scenario). There is no kinetic friction along the table, but there is air friction.

The magnitude of air friction is related to velocity with the equation .2v.
m=3 kg
r=2 m
v(initial)=10 m/s

What is the centripetal acceleration after 5 s.

Homework Equations



F=ma=m(v^2/r)

The Attempt at a Solution



i know you got to integrate the air friction function over the 5 s interval, but i can't express it in terms of t.

I am completely lost, please help. I know I got it wrong, but i really want to know how to do it.

Thanks!
F=mv^2/r is the correct centripetal force formula for the tension in the rope, but the problem is asking for the centripetal acceleration only. You need first, however, to use Newton's 2nd law in the tangential direction, where the tangential acceleration is not constant. Thus,
use F=m(dv/dt), where F is the given air resistance force (presumably in force units of Newtons). You get a first order differential equation you must solve of the form -kv = mdv/dt. Solve for v and don't forget the boundary condition. Is this where you are stuck?
 

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