Answer to Question #162874 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

After a few rotations, the man then extends his arms horizontally while carrying

both test masses in each hand. We can simplify this system by considering a solid

cylinder, a slender rod rotating about its center, and two point particles. The solid

cylinder has a diameter equal to the man’s torso width and mas equal to 55.0 kg.

The slender rod has a length equal to the man’s arm span and mass equal to 15.0

kg. The test masses are placed on both ends of the slender rod.

a. Calculate the moment of inertia of the following:

i. Solid cylinder (I = MR^2/2)

ii. Slender rod (I = ML^2/12)

iii. Two test masses (I = MR^2)

b. What is the total moment of inertia of the system?

c. Using the angular momentum obtained in (1-b) as the initial angular

momentum, calculate the final angular momentum after the arms are

extended.

d. Did the system rotate faster or slower?

Answer 1

(a) (i) Let's calculate the moment of inertia of solid cylinder:

$$I=\dfrac{1}{2}MR^2=\dfrac{1}{2}\cdot55\ kg\cdot(0.2\ m)^2=1.1\ kg\cdot m^2.$$

(ii) Let's calculate the moment of inertia of slender rod:

$$I=\dfrac{1}{12}ML^2=\dfrac{1}{12}\cdot15\ kg\cdot(1.75\ m)^2=3.83\ kg\cdot m^2.$$

(iii) Let's calculate the moment of inertia of two test masses:

$$I=MR^2=6\ kg\cdot(0.875\ m)^2=4.6\ kg\cdot m^2.$$

(b)

$$I_{tot}=1.1\ kg\cdot m^2+3.83\ kg\cdot m^2+4.6\ kg\cdot m^2=9.53\ kg\cdot m^2.$$

(c) According to the law of conservation of angular momentum, we get:

$$L_i=L_f=0.76\ \dfrac{kg\cdot m^2}{s}.$$

(d) Let's find the angular velocity  after the arms are extended:

$$L_i=L_f=I_{tot}\omega_f,$$ $$\omega_f=\dfrac{L_i}{I_{tot}}=\dfrac{0.76\ \dfrac{kg\cdot m^2}{s}}{9.53\ kg\cdot m^2}=0.08\ \dfrac{rad}{s}.$$

Initially, when man rotates with his arms close to his chest, the angular velocity equals $0.5\ \dfrac{rad}{s}.$

Therefore, the system rotates slower.