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When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_x\) and \(a_z\) are approximately zero and \(v_x\) and \(\omega_z\) are approximately constant. Rolling without slipping means \(v_x = r\omega_z\) and \(a_x = r\alpha_z\) . If an object is set in motion on a surface \(without\) these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_0\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_k\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_x\) of the center of mass and \(a_z\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_z = \omega_0\) but \(v_x =\) 0. Rolling without slipping sets in when \(v_x = r\omega_z\) . Calculate the \(distance\) the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

Short Answer

Expert verified
The accelerations are \( a_x = \mu_k g \) and \( \alpha_z = \frac{2\mu_k g}{R} \). The distance is \( s = \frac{1}{2} \mu_k g \left( \frac{\omega_0}{\alpha_z + a_x/R} \right)^2 \). The work done is \( W = \mu_k M g s \).

Step by step solution

01

Draw the Free-Body Diagram

First, we need to identify the forces acting on the cylinder. The forces include gravity acting downward, the normal force acting upward, and the kinetic friction force acting horizontally. Since the object is initially slipping, the kinetic friction force will oppose the direction of motion. As the cylinder starts to roll, this frictional force also creates a torque that causes angular acceleration.
02

Calculate Acceleration of Center of Mass, \( a_x \)

For the solid cylinder, the horizontal acceleration of the center of mass \( a_x \) is caused by the kinetic friction force. Using Newton's second law \( F = ma \), we have:\[F = \mu_k \cdot M \cdot g = M \cdot a_x\]Solving for \( a_x \):\[ a_x = \mu_k \cdot g \]
03

Calculate Angular Acceleration, \( \alpha_z \)

The angular acceleration is produced by the torque due to friction. Calculating the torque \( \tau = F \cdot R \), we have:\[ \tau = \mu_k \cdot M \cdot g \cdot R \]Since \( \tau = I \cdot \alpha \) and for a solid cylinder \( I = \frac{1}{2}MR^2 \), we can solve for angular acceleration \( \alpha_z \):\[ \alpha_z = \frac{2\mu_k \cdot g}{R} \]
04

Determine the Condition for Rolling Without Slipping

For rolling without slipping, \( v_x = R \cdot \omega \). Initially, \( v_x = 0 \) and \( \omega = \omega_0 \). We use kinematics to find the condition:\[ v = a_x t = R (\omega_0 + \alpha_z t) \]Solving for time \( t \):\[ t = \frac{\omega_0}{\alpha_z + a_x/R} \]
05

Calculate the Distance Rolled Before Slipping Stops

Use the kinematics equation \( v^2 = u^2 + 2as \) to find the distance \( s \):\[ v_x = a_x t = \left( \mu_k \cdot g \right) \cdot t \]Substitute \( t = \frac{\omega_0}{\alpha_z + a_x/R} \) into the distance formula:\[ s = \frac{1}{2} a_x t^2 = \frac{1}{2} (\mu_k \cdot g) \cdot \left( \frac{\omega_0}{\alpha_z + a_x/R} \right)^2 \]
06

Calculate the Work Done by Friction Force

The work done by the friction force is given by \( W = F \cdot s \). Since the friction force is \( \mu_k M g \), the work done over distance \( s \):\[ W = \mu_k \cdot M \cdot g \cdot s \]Substitute the expression for \( s \) from Step 5 to complete the calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Friction
Kinetic friction plays a crucial role when an object slips on a surface. It acts as a resistive force opposing the motion of the object. Kinetic friction is given by the equation \( F_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. This force is particularly significant when an object, like a cylinder in our exercise, begins to slide. Initially, the entire contact motion is slipping, which means kinetic friction is the predominant force at play.

As the cylinder transitions towards rolling without slipping, the kinetic friction force not only opposes the sliding motion but also generates torque. This torque induces angular acceleration, which gradually decreases the slipping effect. Once rolling commences without any slipping, the role of kinetic friction diminishes and becomes negligible as the velocities align according to the condition \( v_x = r\omega_z \). During the process from slip to roll, kinetic friction ensures that energy conversion and motion adaptations occur smoothly.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It is a measure of how quickly an object spins up or slows down. When a cylinder experiences a kinetic friction force, it not only affects the linear acceleration but also creates a torque that results in angular acceleration.

To find the angular acceleration \( \alpha_z \), we use the torque equation \( \tau = I \cdot \alpha \). For a solid cylinder, the moment of inertia \( I = \frac{1}{2}MR^2 \). The kinetic friction force creates a torque, calculated as \( \tau = \mu_k \cdot M \cdot g \cdot R \). Plug these values into the torque equation to solve for \( \alpha_z \): \[ \alpha_z = \frac{2\mu_k \cdot g}{R} \]

The angular acceleration continues to affect the cylinder until the condition for rolling without slipping \( v_x = R\omega_z \) is met. At this point, the cylinder rolls smoothly, and the angular acceleration diminishes because there is no further resistance by kinetic friction.
Free-Body Diagram
To analyze the forces acting on an object, drawing a free-body diagram is essential. In the case of our rolling cylinder, the free-body diagram helps us visualize all the forces and their directions.

For the cylinder on a horizontal surface, three primary forces are at play:
  • Gravitational Force: Acts downward, equal to the weight of the cylinder \( Mg \).
  • Normal Force: Acts upward, balancing the gravitational force.
  • Kinetic Friction: Acts horizontally, opposing the direction of slip, and creating a torque that leads to angular acceleration.
By carefully considering these forces in the free-body diagram, we can proceed with calculations like linear acceleration \( a_x \) and angular acceleration \( \alpha_z \). The diagram serves as a guide to ensure all physical quantities and interactions are accounted for during the analysis. Additionally, it illustrates how these forces interrelate, paving the way for understanding more complex motion dynamics.

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Most popular questions from this chapter

A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.40). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks?

A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

A certain gyroscope precesses at a rate of 0.50 rad/s when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is 0.165g, what would be its precession rate?

A block with mass m is revolving with linear speed \(v_1\) in a circle of radius \(r_1\) on a frictionless horizontal surface (see Fig. E10.40). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_2\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_1\) and the radius \(r_1\). (b) Use \(W = \int_{r1}^{r2} \overrightarrow{T} (r) \cdot d\overrightarrow{r}\) to calculate the work done by \(\overrightarrow{T}\) when \(r\) changes from \(r_1\) to \(r_2\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

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