# Rolling ball WITH slipping

## Homework Statement

A rigid ball of mass m = 6kg and radius R = .23 m is sitting on a table. The mass center and geometric center of the ball coincide, but the radius of gyration about its mass center is k = .19 m. The ball is initially at rest. There is a coefficient of friction μ = .435 between the ball and the table. Two forces are then applied to the ball: a force F1 = 55 N to the right at the center of mass, and a force F2 = 20 N to the left at the very top of the ball. Given this, determine the angular acceleration of the ball and the linear acceleration of the ball's center at the instant the forces are first applied.

## Homework Equations

$I=mk^2$ where k is the radius of gyration.
Acceleration point relation for rigid bodies
ƩF=ma
ƩM=Iα

## The Attempt at a Solution

The acceleration at the point of contact must be zero. Because the ball is initially at rest, this means that the acceleration at the center of the ball is simply αR because the ball is a rigid body. Assume that the friction force f, which acts at the point of contact, points to the left.

Next, we use Newton's law in the x and y directions. This gives F1 - F2 - f = ma = mrα and N - mg = 0. Then we take the sum of the moments about the center of the ball, which yields F2R - fR = mk2α. Without making the assumption that f = μN, we solve these three equations for 3 unknowns to find that N = 58.8 newtons, f = -12.2 newtons and α = 34.2 rad/s2.

This is where I get lost, because clearly f≠μN. How do I reconcile this? What assumptions am I making that are incorrect?

ehild
Homework Helper

## Homework Statement

A rigid ball of mass m = 6kg and radius R = .23 m is sitting on a table. The mass center and geometric center of the ball coincide, but the radius of gyration about its mass center is k = .19 m. The ball is initially at rest. There is a coefficient of friction μ = .435 between the ball and the table. Two forces are then applied to the ball: a force F1 = 55 N to the right at the center of mass, and a force F2 = 20 N to the left at the very top of the ball. Given this, determine the angular acceleration of the ball and the linear acceleration of the ball's center at the instant the forces are first applied.

## Homework Equations

$I=mk^2$ where k is the radius of gyration.
Acceleration point relation for rigid bodies
ƩF=ma
ƩM=Iα

## The Attempt at a Solution

The acceleration at the point of contact must be zero.

The title of your thread is "rolling ball with slipping" If the ball slips, the point of contact accelerates, and the rolling condition aCM=αR does not hold.

Because the ball is initially at rest, this means that the acceleration at the center of the ball is simply αR because the ball is a rigid body. Assume that the friction force f, which acts at the point of contact, points to the left.

Next, we use Newton's law in the x and y directions. This gives F1 - F2 - f = ma = mrα and N - mg = 0. Then we take the sum of the moments about the center of the ball, which yields F2R - fR = mk2α. Without making the assumption that f = μN, we solve these three equations for 3 unknowns to find that N = 58.8 newtons, f = -12.2 newtons and α = 34.2 rad/s2.

This is where I get lost, because clearly f≠μN. How do I reconcile this? What assumptions am I making that are incorrect?

You determined the angular acceleration and force of static friction of the rolling ball, that does not slip. Is the static friction equal to μN?

ehild