Finding velocity given radius and coefficient of friction

[Mooshk]

Homework Statement

A box with mass 2kg is on the edge of a circular platform of radius 6.0m. The coefficient of friction between the platform and the box is 0.3. The platform accelerates. Determine the speed when the box slips off the edge.

Homework Equations

Fs = μsN
F⃗ net=ΣF⃗ =ma⃗
a(t) = d(vt)/dt
a(r) = v^2/r
a = [sqrt (ar^2) + (at^2)]

The Attempt at a Solution

I know that this is non-uniform circular motion.
I don't think that what I'm doing is right.

ΣF(t) = -Ff = m(at)
at = - μs(g)
a(t) = - (0.3)(9.8)
a(t) = -2.94 m/s^2

ΣF(r) = m(ar)
ΣF(r) = (mv^2) / r

ΣF(net) = m * [sqrt (ar^2) + (at^2)]
Ff = m * [sqrt (ar^2) + (at^2)]
μs(g) = [sqrt (μs^2(g^2)) + (m^2*v^4) / r^2]
(μg)^2 = (μs^2(g^2)) + (m^2v^4) / r^2
((μg)^2 - μs^2(g^2)) /r^2 = (m^2*v^4)

So if I solve for v, I would take the fourth root of the left side after bringing m^2 over, and that doesn't seem right to me.

 

[Delphi51]

I think you are working too hard. The acceleration is not given, suggesting you ignore F = ma. To my mind, it sounds like the condition for beginning to slip is simply when friction can no longer provide the necessary centripetal force to hold it on:
centripetal force = force of friction
which easily solves for v.

 

[Mooshk]

I think you are working too hard. The acceleration is not given, suggesting you ignore F = ma. To my mind, it sounds like the condition for beginning to slip is simply when friction can no longer provide the necessary centripetal force to hold it on:
centripetal force = force of friction
which easily solves for v.

How would that model non-uniform circular motion? Or does that not matter since the friction force is perpendicular?

 

[Delphi51]

Strictly speaking, it does not apply to non-uniform circular motion where it could be that high tangential acceleration will cause the thing to slip into kinetic friction mode before centrifugal force overcomes the grip. If you have been doing problems like that in class, then do it with this one, too. Note that you will have to consider the combined radial and tangential forces and calculate when that exceeds the friction force.

But the wording suggests to me that the tangential acceleration is so small it can be ignored and slipping will occur in the radial direction.

 

[asteriatic]

I would approach this problem by first drawing a free body diagram to identify all the forces acting on the box. These include the normal force (N), the weight of the box (mg), and the frictional force (Ff). The platform is accelerating, so there is also a net force (Fnet) acting on the box.

Using the equation Ff = μsN, we can calculate the magnitude of the frictional force. In this case, Ff = (0.3)(2kg)(9.8m/s^2) = 5.88N.

Next, we can use the equation Fnet = ma to determine the acceleration of the box. Since the box is slipping off the edge, the net force must be equal to the frictional force, so we have Fnet = Ff = 5.88N. Plugging this into the equation, we get 5.88N = (2kg)a. Solving for a, we get a = 2.94m/s^2.

Now, we can use the equation a = v^2/r to find the speed of the box when it slips off the edge. Rearranging the equation, we get v = √(ar). Plugging in the values for a and r, we get v = √(2.94m/s^2)(6.0m) = 4.28m/s.

Therefore, the speed of the box when it slips off the edge is 4.28m/s. This approach is more direct and uses the appropriate equations for the given problem. It is important to carefully consider the forces and their directions in order to accurately solve the problem.