Answer to Question #162873 in Physics for Vic

Question

Average male mass = 70.0 kg

Average arm span = 1.75 m

Average torso width = 0.400 m

Test masses = 3.00 kg

Initially, the man makes himself rotate with his arms close to his chest. We can

simplify this system to be a solid cylinder whose diameter is equal to the man’s

torso width and whose mass is equal to the total mass of the man and the two test

masses.

a. Calculate the moment of inertia of this solid cylinder. (Recall that the

moment of inertia of a solid cylinder is given by I = MR^2/2.)

b. Suppose that the system rotates with an angular velocity of 0.500 rad/s.

Find the angular momentum of this system.

Answer 1

(a)

$$I=\dfrac{1}{2}MR^2,$$ $$I=\dfrac{1}{2}\cdot(70\ kg+2\cdot3\ kg)\cdot(0.2\ m)^2=1.52\ kg\cdot m^2.$$

(b)

$$L=I\omega=1.52\ kg\cdot m^2\cdot0.5\ \dfrac{rad}{s}=0.76\ \dfrac{kg\cdot m^2}{s}.$$