Answer to Question #162869 in Physics for Irish

Question

We are to revisit the demonstration of a man holding two test masses sitting on a rotating

chair. Remember in the demonstration that the rotation of the man on the chair slows

done when he extends his arms while holding the two masses. We will analyze this

phenomenon using conservation of angular momentum.

We will assume that the man in the demonstration is of average European composition

(the demonstration video seems to be Russian in origin). Thus, the following dimensions

will be used in the calculations.

Average male mass = 70.0 kg

Average arm span = 1.75 m

Average torso width = 0.400 m

Test masses = 3.00 kg

1. 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)

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