A thin rod of length 2.15 m and mass 13.1 kg is rotated at an angular speed of 3.95 rad/s around an axis perpendicular to the rod and through its center of mass. Find the magnitude of the rod's angular momentum.

L = 239.2 J.s

Explanation:

The angular momentum of the rod is given by the following formula:

L = mvr

where,

L = Angular Momentum = ?

m = mass = 13.1 kg

r = radius of circle of rotation = length of rod = 2.15 m

v = linear speed of rod = r(angular speed of rod) = (2.15 m)(3.95 rad/s)

v = 8.5 m/s

Therefore,

L = (13.1 kg)(8.5 m/s)(2.15 m)

L = 239.2 J.s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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