Spinning ball dropped on the ground

[vladimir69]

Homework Statement

A solid ball of mass M and radius R is spinning with angular velocity $\omega_0$ about a horizontal axis. It drops vertically onto a surface where the coefficient of kinetic friction with the ball is $\mu_k$. Find an expression for the final angular velocity once it has achieved pure rolling motion. The other part of the question asks for the time taken until its in pure rolling motion. I could solve that part easily enough once I knew the answer to the first part.

Homework Equations

$$\tau=I\alpha$$
$$L=I\omega$$
$$RF=i\alpha$$
$$L=RMv$$

The Attempt at a Solution

The answer says
$$\omega=\frac{2}{7}\omega_0$$
I found a way of getting that answer by using
$$I\omega_0=I\omega+RMv=I\omega+MR^2\omega$$
and then solving for omega gives the required result. But I can't understand how using that equation yields the right result, so there must be another way. I don't think you can use conservation of energy for this either.

Thanks,

 

[anathema]

but I'm not sure if this is the right approach. Let's break down the problem into smaller parts. First, we know that the ball is initially spinning with angular velocity \omega_0. This means that it has angular momentum L=I\omega_0. When the ball drops onto the surface, it will start to experience a frictional force, which will cause it to rotate and also move forward. However, we know that the frictional force will eventually bring the ball to a stop, and it will start to roll without slipping. This means that the final angular velocity \omega_f will be related to the final linear velocity v_f by the equation \omega_f=\frac{v_f}{R}.

Now, let's consider the forces acting on the ball while it is in motion. We have the frictional force F_f acting in the opposite direction of the motion, and we also have the weight of the ball mg acting downwards. Since the ball is in pure rolling motion, we can also say that the net torque acting on the ball is zero. This means that the sum of the torques due to the frictional force and the weight must equal zero.

Since we know that \tau=F_fR and RF=i\alpha, we can rewrite this equation as F_fR=i\alpha. We also know that the moment of inertia for a solid ball is I=\frac{2}{5}MR^2. Substituting this into the equation, we get F_fR=\frac{2}{5}MR^2\alpha.

Now, we can use Newton's second law for rotation, \tau=I\alpha, to find an expression for the acceleration \alpha. Substituting in the value for the moment of inertia, we get F_fR=\frac{2}{5}MR^2\frac{\tau}{I}. Since we know that \tau=mgR, we can rewrite this as F_fR=\frac{2}{5}MR^2\frac{mgR}{I}.

Next, we can use Newton's second law for linear motion, F=ma, to find an expression for the acceleration a. Substituting in the value for the frictional force, we get \mu_kmg=\frac{2}{5}MRa. Solving for a, we get a=\frac{5}{2}\frac{\mu_kg}{M}.

Finally,

 

[nlightner]

but I'm not sure what answer you are referring to. Can you clarify?

In general, when a spinning ball is dropped onto a surface, it will experience a frictional force that will cause it to slow down and eventually reach a state of pure rolling motion. This is because the frictional force will act to both slow down the rotation and to accelerate the ball forward in the direction of its motion.

To find the final angular velocity of the ball once it has achieved pure rolling motion, we can use the equation for torque, \tau=I\alpha, where \tau is the torque acting on the ball, I is the moment of inertia, and \alpha is the angular acceleration. In this case, the torque is due to the frictional force, which can be calculated as F_f=\mu_k mg, where \mu_k is the coefficient of kinetic friction, m is the mass of the ball, and g is the acceleration due to gravity.

We also know that the angular velocity of the ball can be related to its linear velocity by the equation L=I\omega=RMv, where L is the angular momentum, R is the radius of the ball, and v is the linear velocity.

Combining these equations, we can set the torque equal to the product of the ball's moment of inertia and angular acceleration, and solve for \alpha:

\tau=I\alpha
\mu_k mg = I\alpha
\alpha = \frac{\mu_k mg}{I}

To find the final angular velocity, we can use the equation for linear velocity, L=RMv, and substitute in the value for \alpha that we just found:

L=I\omega=RMv
\omega = \frac{L}{I} = \frac{RMv}{I} = \frac{RM}{I} \left(\frac{\mu_k mg}{I}\right)

Finally, we can use the parallel axis theorem to substitute in the moment of inertia for a solid sphere, which is I=\frac{2}{5}MR^2:

\omega = \frac{RM}{I} \left(\frac{\mu_k mg}{I}\right) = \frac{2}{5} \frac{MR}{\frac{2}{5}MR^2} \left(\frac{\mu_k mg}{\frac{2}{5}MR^2}\right) = \frac{5}{2} \