# Rolling Motion with slipping

• jbphys303
In summary, a uniform density sphere with an angular speed of 10 rev/sec and no initial linear velocity slips along the surface before pure rolling without slipping begins. The coefficient of friction between the sphere and the surface is 0.21, and the radius and mass of the sphere are 0.08m and 7.3 kg, respectively. To determine the speed of the ball at time t, the equations for linear and angular acceleration can be used, along with the fact that at the time when v=rw, static friction takes place and the ball begins to roll. The time it takes to reach this speed can then be found, and the amount of energy lost by friction between t=0 and t=t can also be calculated
jbphys303

## Homework Statement

A uniform density sphere is released such that it has an angular speed of 10 rev/sec and no initial linear velocity. The angular velocity vector is perfectly perpendicular to the linear momentum vector. Initially the ball slips as it moves along the surface, but after time t pure rolling without slipping begins.

The coefficient of friction between the sphere and the surface is 0.21.
The radius of the ball is 0.08m
mass sphere= 7.3 kg
w= 10 rev/sec

#1: How fast is the ball rolling at time t?

#2: How long did it take to reach this speed?

#3: How much energy was lost between time t=0 and t=t?

## Homework Equations

Rotational Inertia (I)=2/5mr^2
kinetic energy (rotational)= 1/2Iw^2
ke= 1/2mv^2
v=rw

## The Attempt at a Solution

I think I can figure it out once I know how much energy is lost by friction
So far I set up the following equation
1/2Iw^2-(energy lost by friction)= 1/2Iw(final)^2+ 1/2mv^2
Any help would be appreciated

Last edited:
The motion of the ball is composed of translation of its CM and of rotation around the CM. The kinetic friction points forward, accelerates translation and its torque with respect to the CM decelerates rotation. At the time when v=rw, static friction takes place and the ball rolls.

Use both equation for linear acceleration ma = F and for angular acceleration I*dw/dt=-RF. Solve for v and w. Find the time when wR=v.

ehild

I would approach this problem by first analyzing the given information and identifying any relevant equations and principles.

From the given information, we know that the sphere has an initial angular speed of 10 rev/sec and no initial linear velocity. This means that the sphere is rotating but not moving in a linear direction. We also know that the angular velocity vector is perpendicular to the linear momentum vector, which is a result of the sphere slipping along the surface.

At a certain time t, pure rolling without slipping begins, which means that the sphere is now both rotating and moving in a linear direction without any slipping. This indicates that the friction between the sphere and the surface has increased to the point where it can prevent slipping.

To answer the questions, we can use the equations for rotational inertia, kinetic energy, and velocity. We know the mass and radius of the sphere, so we can calculate its rotational inertia using the equation I = 2/5 * m * r^2.

To answer question #1, we can use the equation v = r * w, where v is the linear velocity, r is the radius, and w is the angular velocity. This equation relates the linear and angular velocities for an object in pure rolling motion. Therefore, we can calculate the linear velocity of the sphere at time t.

For question #2, we can use the equation for kinetic energy, K = 1/2 * I * w^2, where K is the kinetic energy, I is the rotational inertia, and w is the angular velocity. We know the initial and final angular velocities, so we can calculate the change in kinetic energy between time t=0 and t=t. This change in kinetic energy is equal to the energy lost by friction, as stated in the problem.

Finally, for question #3, we can simply calculate the energy lost by friction by using the previously calculated change in kinetic energy.

Overall, to solve this problem, we need to use the equations for rotational inertia, kinetic energy, and velocity, and take into account the information about pure rolling without slipping and the coefficient of friction.

## 1. What is rolling motion with slipping?

Rolling motion with slipping is a type of motion where an object is both rotating and translating at the same time. This typically occurs when an object is rolling on a surface with some amount of friction, causing it to slip while rolling.

## 2. What causes rolling motion with slipping?

Rolling motion with slipping is caused by a combination of rotational and translational forces acting on an object. The friction between the object and the surface it is rolling on can cause it to slip while rolling, leading to this type of motion.

## 3. How is rolling motion with slipping different from rolling without slipping?

In rolling without slipping, there is no slipping or sliding between the object and the surface it is rolling on. The point of contact between the object and the surface remains stationary. However, in rolling motion with slipping, the point of contact is not stationary, and there is some degree of slipping between the object and the surface.

## 4. Can rolling motion with slipping be controlled or prevented?

In most cases, rolling motion with slipping cannot be completely controlled or prevented. However, factors such as the surface material, the shape and size of the object, and the amount of force applied can affect the degree of slipping. In some cases, adding additional friction, such as using a rougher surface, can minimize slipping.

## 5. How is rolling motion with slipping relevant in real-world applications?

Rolling motion with slipping is relevant in many real-world applications, such as car tires on a road, balls rolling on a sports field, or wheels on a skateboard. Understanding this type of motion is important in designing and optimizing these objects for their intended use and ensuring safety. It is also relevant in physics and engineering studies for analyzing and predicting the behavior of rolling objects.

• Introductory Physics Homework Help
Replies
35
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
534
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
97
Views
3K
• Introductory Physics Homework Help
Replies
39
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
753
• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
1K