Question

A 0.25 ball is thrown with the speed of v=4.5m/s at a resting stick that can rotate around one end at the axis O. The ball hits the stick at r=0.75 m from O. What is the angular momentum of the ball about the axis O before hitting the stick ?

Answer 1

Answer:

$\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^{-1}}$

Explanation:

mass of the ball = 0.25  kg

its speed v = 4.5 m/s  

r = 0.75 m

let the speed be in the horizontal direction and the distance r be in the vertical direction we have :

$^ {\to}v = 4.5 \ x \ \ m/s$

$^ {\to}r = 0.75 \ y \ \ m/s$

Let the momentum about the center of intersection be P;

SO;

$^ \to} P = ^{\to} mv$

$^ \to} P = 0.25* 4.5 \ x \ \ m/s$

$^ \to} P = 1.125 \ x \ \ m/s$

Let the angular momentum be Q;

$^\to Q = ^ \to r* ^ \to P$

$^\to Q = (0.75*1.125) kgm^2s^{-1} *(\hat y * \hat x)$

$\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^{-1}}$