Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. assume the clock hand has a length of 15.0 cm and a mass of 6.00 g. consider the second hand to be a slender rod rotating with constant angular velocity about one end. (express your answer in the form: "1.23e-4 kg*m^2/s'.)

L = 4.711 *10^{-6} kg m2/s

Explanation:

   =4.5*10^-5

angular velocity

             = 0.1047 rad/s

the angular momentum,

L=4.711×

Explanation:

we have given mass m=6 gram =0.006 kg

length =15 cm =0.15 m

moment of inertia for the clock hand is given by

angular velocity

we know that angular momentum is given by

L=Iω

L=4.5××0.1047

L=4.711×

the answer is b i think