Question

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4.5 kg⋅m2 . A second student tosses a 2.0 kg mass with a speed of 2.5 m/s to the student on the stool, who catches it at a distance of 0.35 m from the axis of rotation. What is the resulting angular speed of the student and the stool?

Answer 1

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

$mvL = (I + mL^2)\omega$

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

$2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega$

$1.75 = [4.5 + 0.245]\omega$

$1.75 = 4.745\omega$

$\omega = 0.37 rad/s$

so the final angular speed will be 0.37 rad/s