Question

A scientist makes a device to catch baseballs. A long bar of total mass 2.2kg and length 1.2m is fixed at its center. It catches a ball of mass 0.15kg moving at v=16m/s in a box at one of its ends. Ignore the mass of the catcher box for this problem. What is the angular velocity of the system after the ball is caught?

Answer 1

Answer:

ωf = 4.53 rad/s

Explanation:

By conservation of the angular momentum:

Ib*ωb = (Ib + Ic)*ωf

Where

Ib is the inertia of the ball

ωb is the initial angular velocity of the ball

Ic is the inertia of the catcher

ωf is the final angular velocity of the system

We need to calculate first Ib, Ic, ωb:

$Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2$

$Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2$

ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s

Now, ωf will be:

$\omega f = \frac{Ib*\omega b}{Ib + Ic} = 4.53rad/s$