What is the angular momentum of the second hand on a clock if the clock has a length of 25cm, a mass of 20g, and is rotating with constant angular velocity. Treat the second hand as a thin rod.

I w

where I is moment of inertia
and
w is radians/second = 2 pi/period
period = 1 minute = 60 seconds

I = (1/3) M L^2 = (1/3)(.020 Kg)(.25 m)^2
= 4.17*10^-4
4.17 *10^-4 * 2 * pi / 60 = 4.36 * 10^-5

To calculate the angular momentum of the second hand on a clock, we need to use the formula for angular momentum:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

To calculate the moment of inertia of the second hand, we can treat it as a thin rod rotating around one end. The moment of inertia of a thin rod rotating around one end is given by the formula:

I = (1/3) * m * L^2

where m is the mass of the rod and L is its length.

Given that the length of the second hand is 25 cm (or 0.25 m) and the mass is 20 g (or 0.02 kg), we can substitute these values into the moment of inertia formula.

I = (1/3) * 0.02 kg * (0.25 m)^2

I = 8.33 * 10^-4 kg * m^2

Now, we need to determine the angular velocity. The second hand completes one full rotation in 60 seconds. Since the angular velocity is constant, we can calculate it by dividing the angle of rotation (2π rad) by the time taken (60 s):

ω = (2π rad) / (60 s)

Now, we can substitute the calculated moment of inertia and angular velocity into the angular momentum formula:

L = (8.33 * 10^-4 kg * m^2) * [(2π rad) / (60 s)]

Simplifying the expression, we can find the value of L, which represents the angular momentum of the second hand on the clock.