A thin rod of length 3.05 m and mass 15.7 kg is rotated at an angular speed of 3.60 rad/s around an axis perpendicular to the rod and through one of its ends. Find the magnitude of the rod's angular momentum.

526 kg-m2/sec

Explanation:

The radius = length of the rod= 3.05 m

angular velocity = 3.60 rad/s

mass = 15.7 kg

L = I ω

Where;

L = angular momentum

I = moment of inertia = mr^2

ω = angular velocity

substituting;

L= 15.7 * (3.05)^2 * 3.60

L = 526 kg-m2/sec

0.11 m/s and 0.28 m/s should be correct

f is prop to m1xm2 double m1 double force