a disc mass m radius r rolls down inclned planemaking angle30 static,kinetic friction=u then friction force=?

Question

inclined plane is rough

Answer 1

Ok, you need to use torques, and the moment of inertia of the disc.

The moment of inertia (I) is = M*R^2/2

You know that there is a friction force : f

And there is a force that is making the disc rolls : mgsen(30)

α : angular acceleration

Now, using torques :

I.α = r*f

r*f = M*r^2*α / 2 ....(1)

Now, the disc is moving to the end :

m*a = mgsen(30) - f....

now : m(α*r) = 1/2*mg - f ....(2)

Using one and two :

α = 2*f / M*r

and : M(2*f/ M*r)*r = 1/2*mg - f

now : 2f = 1/2*mg - f

3f = 1/2mg

f = 1/6*mg

Hope that might help you. I use the rotation dinamics, that's all.

Answer 2

KE in the oposit diretion = 1/2 mv^2
F=ma
resolve for the angle baby, thats 4 u
the rest is history

cheers