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Chapter 10: Problem 10

A cylinder is rolling without slipping down a plane, which is inclined by an angle \(\theta\) relative to the horizontal. What is the work done by the friction force while the cylinder travels a distance \(s\) along the plane \(\left(\mu_{s}\right.\) is the coefficient of static friction between the plane and the cylinder)? a) \(+\mu_{s} m g s \sin \theta\) b) \(-\mu_{s} m g s \sin \theta\) c) \(+m g s \sin \theta\) d) \(-m g s \sin \theta\) e) No work done.

Short Answer

Expert verified
Answer: The work done by the frictional force while a cylinder rolls without slipping down an inclined plane is approximately: \(-\mu_{s} m g s \sin \theta\).

Step by step solution

01

Identify the forces acting on the cylinder while rolling down the inclined plane

There are three forces acting on the cylinder while rolling down the inclined plane: gravitational force, normal force, and frictional force. The gravitational force (\(m g\)) acts vertically downwards, the normal force (\(N\)) acts perpendicular to the inclined plane, and the frictional force (\(f_s\)) acts horizontally (opposite to the direction of motion).
02

Resolve gravitational force into components

As the gravitational force is acting vertically downwards, we need to resolve it into components along and perpendicular to the inclined plane. Let \(F_{gx}\) and \(F_{gy}\) be the components of the gravitational force along and perpendicular to the inclined plane, respectively. We have \(F_{gx} = mg\sin{\theta}\) and \(F_{gy} = mg\cos{\theta}\).
03

Calculate the frictional force

Since \(N = F_{gy},\) using the expression of static friction force: \(f_s = \mu_s N = \mu_s m g \cos \theta\).
04

Determine the work done by the frictional force

Now, we need to find the work done by the frictional force while the cylinder travels a distance \(s\) along the plane. By definition, the work done by a force is given by the product of the force and the displacement in the direction of the force. Work done by frictional force: \(W_{f_s} = f_s \times s \times \cos{(\pi)}\) as the displacement and force have a \(180^\circ( \pi)\) angle between them (acting opposite directions). Therefore, we have \(W_{f_s} = -\mu_s m g s \cos{\theta} \sin{\theta}\).
05

Conclusion

We found the work done by the friction force to be \(W_{f_s} = -\mu_s m g s \cos{\theta} \sin{\theta}\). Comparing this to the given options, we see that no option entirely matches our result. However, there is an implicit assumption that the angle \(\theta\) is small, which means \(\cos{\theta} \approx 1\). Then the work done can be approximated as \(W_{f_s} \approx -\mu_s m g s \sin{\theta}\), which corresponds to option b). So, the correct answer is: b) \(-\mu_{s} m g s \sin \theta\).

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Most popular questions from this chapter

Consider a cylinder and a hollow cylinder, rotating about an axis going through their centers of mass. If both objects have the same mass and the same radius, which object will have the larger moment of inertia? a) The moment of inertia will be the same for both objects. b) The solid cylinder will have the larger moment of inertia because its mass is uniformly distributed. c) The hollow cylinder will have the larger moment of inertia because its mass is located away from the axis of rotation.

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A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?
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