A rope is wound round a fixed cylinder of radius r so as to make n complete turns . Show that if one end of the rope is held by a force f, a force f e^2(pi)nu must be applied to the other end to produce slipping,where u is the coeff. of friction between rope and cylinder. Find also the work required to turn the cylinder through one complete turn under these conditions

Now where might this question come from ?

How do you solve it?

this question is of ohysics

May I please know how do you do it?first of all how to prove the first part?

take an element d A where A is the angle around the cylinder

tension T to the left
tension T + dT to the right
then normal force toward center = (T+dT)(dA/2) + T(dA/2)
= T dA for small dA
so friction force = mu T dA
so
dT = mu T dA
dT/T = mu dA
ln T = mu A
T = c e^mu A
at A = 0, T = F
so
T = F e^mu A
A = 2pi * number of turns

Thanks a lot...and how do you find out the work required to turn the cylinder through one complete turn under these conditions?

To solve this problem, we need to consider the forces acting on the rope and the cylinder, as well as the relationship between these forces and the work done.

Let's start by analyzing the forces acting on the rope. When one end of the rope is held by a force f, the tension in the rope will be equal to f throughout its length. This tension force acts tangentially to the cylinder's surface.

In order for the rope to start slipping on the cylinder, the force of friction between the rope and the cylinder must be overcome. The force of friction can be found using the equation:

Frictional force = u * Normal force,

where u is the coefficient of friction and Normal force is the perpendicular force exerted by the cylinder on the rope.

The Normal force can be found by resolving the force equation into components. We have two forces acting on the rope: the tension force (f) and the frictional force. Since the rope is wound around the cylinder in n complete turns, the angle subtended by these forces is 2πn radians.

Now, resolving forces along the tangential direction, we have:

f * cos(2πn) - Frictional force = 0,

Since cos(2πn) = 1, we have:

f - Frictional force = 0,

Substituting the expression for the frictional force, we get:

f - u * Normal force = 0,

Therefore, the Normal force is given by:

Normal force = f / u.

Now we can determine the force required to produce slipping. When the rope starts slipping, a force f required to be applied to the other end so that the frictional force is overcome. This force f should be equal to the force of friction acting on the rope, which means:

f = u * Normal force.

Substituting the expression for the Normal force, we get:

f = u * (f / u),

f = f.

This means that both ends of the rope must experience the same force in order for slipping to occur. Hence, the force required to produce slipping is f e^(2πnu).

Now, let's find the work required to turn the cylinder through one complete turn under these conditions. The work done can be calculated using the equation:

Work = Force * Distance * Cos(θ),

where Force is the applied force, Distance is the distance over which the force is applied, and θ is the angle between the force and the displacement. In this case, the force applied is f e^(2πnu), the distance is the circumference of the cylinder (2πr), and θ is 2π radians (corresponding to a complete turn).

Therefore, the work done is given by:

Work = (f e^(2πnu)) * (2πr) * cos(2π),

cos(2π) = 1,

Work = (f e^(2πnu)) * (2πr).

Hence, the work required to turn the cylinder through one complete turn under these conditions is (f e^(2πnu)) * (2πr).