Question

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'.)

Answer 1

Answer:

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

Explanation:

$i =\frac{ml^3}{3}$

$= \frac{0.00*.15^2}{3}$

   =4.5*10^-5

angular velocity

$\omega = \frac{2\pi}{60}$

             = 0.1047 rad/s

the angular momentum,

$L = I\omega$

$L = 4.5*10^{-5}* 0.1047 rad/s$

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

Answer 2

Answer:

L=4.711×$10^{-6}$$Kgm^2/sec$

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

$I=\frac{ml^2}{3}$

$I=\frac{0.006\times .15^2}{3}=4.5\times 10^{-5}$

angular velocity $\omega =\frac{2\times \pi}{60}=0.1047 rad/sec$

we know that angular momentum is given by

L=Iω

L=4.5×$10^{-5}$×0.1047

L=4.711×$10^{-6}$$Kgm^2/sec$