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.
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.