Finding velocity given radius and coefficient of friction

AI Thread Summary
To determine the speed at which a 2kg box slips off a circular platform with a radius of 6.0m and a coefficient of friction of 0.3, one must analyze the forces acting on the box. The key point is that slipping occurs when the frictional force can no longer provide the necessary centripetal force. The discussion emphasizes that the acceleration is not provided, suggesting that the focus should be on the balance between centripetal force and friction. While some participants argue about the relevance of non-uniform circular motion, the consensus leans towards simplifying the problem by treating tangential acceleration as negligible. Ultimately, the solution hinges on equating the centripetal force to the frictional force to find the critical speed.
Mooshk
Messages
5
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
Delphi51 said:
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?
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top