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.

Question

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.

Answer 1

http://www.jiskha.com/display.cgi?id=1394740694#1394740694.1394743546

Answer 2

I did not understand the part of the normal force..how did you gt that?

Answer 3

To solve this problem, let's start by analyzing the forces acting on the rope. Consider a small segment of the rope, of length Δs, which subtends an angle Δθ at the center of the cylinder.

1. Force Required to Overcome Friction:
The force required to overcome friction between the rope and the cylinder can be determined by considering the frictional force acting on this small segment of the rope. The frictional force is given by:

f_friction = u * Normal force

The normal force is the force applied perpendicular to the surface, which is balancing the component of force applied parallel to the surface (tangential to the cylinder) on the small segment of the rope.

Since the rope is wound around the cylinder, the normal force is equal to the tangential force applied on the small segment. Hence, the normal force for this small segment is:

f_normal = Δs * f / (2πr)

Therefore, the frictional force on this small segment is:

f_friction = u * (Δs * f / (2πr))

2. Work Required to Turn the Cylinder:
The work required to turn the cylinder through one complete turn can be found by integrating the work done by each small segment of the rope over the entire length of the rope.

The work done by one small segment of the rope is given by:

dW = (f_friction) * (Δs) * cos(Δθ)

Integrating this expression over the entire length of the rope, from 0 to 2πnu, where n is the number of turns and u is the coefficient of friction, we get:

W = ∫[0,2πnu] (f_friction) * (Δs) * cos(Δθ)

Substituting the expression for f_friction and simplifying:

W = u * f * ∫[0,2πnu] (Δs) * cos(Δθ) / (2πr)

Since cos(Δθ) = √(1 - sin²(Δθ)), and Δs = r * Δθ, we can further simplify:

W = u * f * r * ∫[0,2πnu] Δθ * cos(Δθ) / (2πr)
W = (u * f * r) / 2 * ∫[0,2πnu] Δθ * cos(Δθ)

Now, let's analyze the first part of the problem statement:

3. Show that force = f * e^(2πnu):
To produce slipping, the force applied to the other end of the rope must be equal to the frictional force on the small segment. Summing up the frictional forces on all the small segments gives us the total force applied on the rope.

f_total = ∫[0,2πnu] f_friction dθ
= u * f * ∫[0,2πnu] Δs * cos(Δθ) / (2πr)
= (u * f * r) / 2 * ∫[0,2πnu] Δθ * cos(Δθ)

Now, we need to calculate this integral to simplify further.

Integrating Δθ * cos(Δθ) from 0 to 2πnu can be done using integration by parts. By applying integration by parts twice, we get the integral as:

∫[0,2πnu] Δθ * cos(Δθ) dθ = 2πnu

Substituting this result back into the total force equation:

f_total = (u * f * r) / 2 * 2πnu
= u * f * r * πnu
= f * e^(2πnu)

Thus, the force required to produce slipping on the other end of the rope is f * e^(2πnu).

To summarize:
- The force required to produce slipping on the other end of the rope is f * e^(2πnu).
- The work required to turn the cylinder through one complete turn under these conditions is (u * f * r * πnu).