A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4.2 . A second student tosses a 1.1 mass with a speed of 3.0 to the student on the stool, who catches it at a distance of 0.45 from the axis of rotation. What is the final rotational speed of the stool, student and mass?

the I for the added mass is mr^2. The w for the added mass is v/r

so do this, add the two momentums and set them equal to (Itotal)Wfinal
and solve for wfinal

To determine the final rotational speed of the stool, student, and mass, we can apply the principle of conservation of angular momentum.

The angular momentum of an object is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Let's denote the moment of inertia of the student-stool system as I1, the final moment of inertia of the system after catching the mass as I2, and the initial angular velocity of the system as ω1.

Before the mass is caught, the total angular momentum of the system is given by:

L1 = I1 * ω1

After the mass is caught, the total angular momentum of the system is given by:

L2 = (I1 + m * r^2) * ω2

where m is the mass of the tossed object and r is the distance at which it is caught.

Since the angular momentum is conserved, we have:

L1 = L2

I1 * ω1 = (I1 + m * r^2) * ω2

Substituting the given values:

4.2 * ω1 = (4.2 + (1.1 * 0.45^2)) * ω2

Simplifying:

4.2 * ω1 = (4.2 + 0.279) * ω2

4.2 * ω1 = 4.479 * ω2

Now we can solve for ω2, the final angular velocity:

ω2 = (4.2 * ω1) / 4.479

Let's substitute the value of ω1:

ω2 = (4.2 * ω1) / 4.479

Finally, we can calculate the final rotational speed by multiplying the angular velocity by the radius of rotation. Assuming the radius of rotation is the same distance at which the mass is caught (0.45 m), we get the final rotational speed:

v = ω2 * r

v = [(4.2 * ω1) / 4.479] * 0.45

You need to substitute the value of ω1 into this equation to get the final rotational speed.