A Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end. Express your answer in kilogram-meters squared per second.

L = 1.41 x 10⁻⁵ kg.m²/s

Explanation:

First, we need to find the linear speed of the second hand of the clock. Since, we know that second hand completes one revolution in 60 seconds. Therefore, its angular velocity will be:

ω = (1 rev/60 s)(2π rad/ 1 rev)

ω = 0.105 rad/s

Now, for linear speed we use formula:

v = rω

where,

v = linear speed = ?

r = radius of circle = length of hand = 15 cm = 0.15 m

Therefore,

v = (0.15 m)(0.105 rad/s)

v = 0.016 m/s

Now, to find the angular momentum of second hand, we use the following formula:

L = m v r

where,

L = Angular Momentum of Second-Hand = ?

m = mass of second-hand = 6 g = 0.006 kg

Therefore,

L = (0.006 kg)(0.016 m/s)(0.15 m)

L = 1.41 x 10⁻⁵ kg.m²/s

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