Problem 101 - University Physics with Modern Physics

Question

When an object is rolling without slipping, the rolling friction force is much less 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 \(\alpha_{z}\) 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 speed \(\omega_{0}\) about the kinetic friction coefficient is \(\mu_{\mathrm{k}}\) (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 \(\alpha_{z}\) 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.

Step-by-Step Solution

Answer

The cylinder rolls a distance of \( \frac{R^2 \omega_0^2}{18\mu_k g} \) before slipping stops, and the work done by friction is \( \frac{MR^2 \omega_0^2}{18} \).

Step 1: Understand the Problem

We have a solid cylinder initially rotating with an angular speed \( \omega_0 \) and dropped onto a surface with an initial velocity of zero. Given rolling without slipping happens when \( v_x = R \omega_z \), we need to find the accelerations, the distance rolled before slipping stops, and work done by the friction force.

Step 2: Drawing the Free-Body Diagram

For the solid cylinder, the forces involved are the gravitational force \( Mg \) acting downward, the normal force \( N \) acting upward, and the kinetic friction force \( f_k = \mu_k N \) acting opposite to the direction of initial motion. Since the cylinder is set down on the surface, \( N = Mg \). The kinetic friction force will act backwards, causing both a linear acceleration \( a_x \) and an angular acceleration \( \alpha_z \).

Step 3: Calculate Accelerations

Using Newton's second law, the linear acceleration \( a_x \) is determined by the equation \( f_k = Ma_x \). Since \( f_k = \mu_k Mg \), we have:\[ a_x = \mu_k g \]For angular acceleration \( \alpha_z \), using the relationship between torque and angular acceleration, \( f_k \cdot R = I \alpha_z \) where \( I = \frac{1}{2}MR^2 \) for a solid cylinder:\[ \mu_k MgR = \frac{1}{2}MR^2 \alpha_z \]Solving for \( \alpha_z \), we get:\[ \alpha_z = \frac{2\mu_k g}{R} \]

Step 4: Determine Distance Rolled Before Slipping Stops

The condition \( v_x = R \omega_z \) gives the point where rolling without slipping starts. Initially, \( v_x = 0 \) and \( \omega_z = \omega_0 \). The equations of motion are:\[ v_x = a_x t \]\[ \omega_z = \omega_0 - \alpha_z t \]Setting \( v_x = R \omega_z \), we substitute the expressions above:\[ \mu_k gt = R(\omega_0 - \frac{2\mu_k g}{R}t) \]Solving for \( t \), we find:\[ t = \frac{R\omega_0}{3\mu_k g} \]The distance \( d \) before sliding stops is:\[ d = \frac{1}{2} a_x t^2 = \frac{1}{2} \mu_k g \left(\frac{R\omega_0}{3\mu_k g}\right)^2 = \frac{R^2 \omega_0^2}{18\mu_k g} \]

Step 5: Calculate the Work Done by Friction Force

The work done by the friction force \( f_k \) is given by \( W = f_k d \). Since \( f_k = \mu_k Mg \) and \( d \) is calculated in Step 4:\[ W = \mu_k Mg \cdot \frac{R^2 \omega_0^2}{18\mu_k g} = \frac{MR^2 \omega_0^2}{18} \]

Key Concepts

Kinetic Friction

When an object such as a cylinder is moving over a surface, kinetic friction comes into play. This is the force that opposes the motion between the moving object and the surface. Kinetic friction differs from static friction, which acts when the object is stationary. In our exercise, the kinetic friction force, denoted as \( f_k \), is calculated using the equation \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. For a cylinder placed on a flat surface, the normal force is simply the weight of the cylinder, \( Mg \), where \( M \) is the mass and \( g \) is the acceleration due to gravity.

However, kinetic friction not only acts against the direction of linear motion but also causes rotational effects. This is because the point of contact tries to slide against the surface, producing both translational and rotational motion of the cylinder. It's important to remember that kinetic friction is smaller than static friction, which allows objects to roll more easily once they start slipping.

Angular Acceleration

Angular acceleration describes how quickly an object's rotational speed changes. For our rolling cylinder, the angular acceleration, \( \alpha_z \), is influenced by the kinetic friction force acting as a torque. Torque is the rotational analog of force, and it causes changes in rotational motion.

The torque \( \tau \) applied by the kinetic friction is given by \( \tau = f_k \cdot R \), where \( R \) is the radius of the cylinder. Since torque also relates to angular acceleration through the cylinder's moment of inertia \( I \), we have \( \tau = I \alpha_z \). For a solid cylinder, the moment of inertia is \( I = \frac{1}{2}MR^2 \). By equating these expressions, we derive the angular acceleration, \( \alpha_z = \frac{2\mu_k g}{R} \).

This means that the cylinder's change in rotational speed is directly proportional to the coefficient of kinetic friction and inversely proportional to its radius. A larger radius implies a slower angular acceleration for the same frictional force.

Torque

Torque is a measure of how much a force acting on an object causes it to rotate. It is often called the "rotational equivalent of force." In our exercise, the kinetic friction force generates a torque on the rolling cylinder. Torque is essential in determining the rotational motion of the object, just like force is essential for linear motion.

The direction of the torque depends on the direction of the force and the position where it is applied. For the rolling cylinder, the torque due to kinetic friction is calculated as \( \tau = f_k \cdot R \). It is this torque that results in the angular acceleration of the cylinder.

In the context of our problem, torque informs the change in angular velocity of the cylinder from its initial rotational state to the rolling condition. Understanding torque helps in predicting how objects will rotate when different forces are applied, and it underscores the fundamental connection between translational and rotational dynamics.