Question

Three small balls of mass 5.6 kg, 2.2 kg, and 2.8 kg are connected by light rods laying along the y-axis. The rod connecting the first and second balls is 4.6 m long and the rod connecting the second and third balls is 2.5 m. The entire system rotates around the x-axis, which is between the first and second balls and a distance 2.9 m from the first ball, at a rotational speed 1.2 s^-1

(a) What are the linear speeds of the three objects?
First Object:  3.48 m/s
Second Object:  2.04 m/s
Third Object: 5.04 m/s
(b) What is the linear kinetic energy of each of the balls?
First Object:  33.909 J
Second Object:  4.577 J
Third Object:  35.56224 J
(c) What is the total kinetic energy of the system? 74.04824 J
(d) What is the moment of inertia around the x-axis of each of the balls?
First Object:  47.096 kg*m^2
Second Object:  6.358 kg*m^2
Third Object:  49.392 kg*m^2
(e) What is the rotational kinetic energy of each of the balls?
First Object:  33.909 J
Second Object:  4.5776 J
Third Object:  35.56224 J
(f) What is the rotational kinetic energy of the system?
74.049 J
(g) What is the distance of the center of mass of the system from the first ball?
(h) What moment of inertia of each ball around the center of mass?
First Object:
Second Object:
Third Object:
(i) What is the moment of inertia of the system with respect to the center of mass? It's not 2.97, 3.4188, or 2.31
(j) What is the moment of inertia due to the shift of the rotational axis from the center of mass to the x-axis?
(k) What is the moment of inertia of the system around the x-axis?
(l) What is the linear speed of the system?
(m) What is the rotational speed of the system?
(n) What is the linear kinetic energy of the system?
(o) What is the rotational kinetic energy of the system? 74.049 J

don't know g-n help please and thank you!

Answer 1

y co-ordinates of the balls

y₁ = -2.9 ; y₂ = 4.6-2.9 = 1.7 ; y3 = 1.7 +2.5 = 4.2

m1 = 5.6 kg ; m2 = 2.2 kg ; m3 = 2.8 kg

g) COM = m_ir_i / m_i  

COM of the system

Y_com = (5.6*-2.9 + 2.2*1.7 + 2.8*4.2) /(5.6+2.2+2.8)

= -0.0698

distance of COM to first ball = -0.0698 +2.9 = 2.83 m

h) MI of the first ball about COM

I_1cm = mr² = 5.6 * 2.83² = 44.85 kg-m²

distance of second ball from COM = -0.0698 - 1.7 = 1.77 m

MI of the second ball about COM I_2cm = 2.2 * 1.77² = 6.89 kg-m²

distance of third ball = 4.2 = 0.0698 = 4.27 m

I_3cm = 2.8 *4.27² = 51.05 kg-m²

i) MI of the system about COM

I_cm = I_1cm + I_2cm + I_3cm = 44.85 + 6.89 + 51.05 = 102.79 kg-m²

j) MI due to shift = Md² = 10.6 * 0.0698 = 0.74 kg-m²

k) MI about x-axis ( use parallel axis theorem )

= I_cm + Md² ( d - shift of axis)

= 102.79 + 10.6 * 0.0698² = 103.53 kg-m²

l) The system as whole has no linear motion , it has only rotational motion.

linear speed of the system = 0

m)

rotational speed = = 1.2 rads/s

n) linear KE =0 , no linear motion

o) RKE = 1/2 * I² = 103.53 * 1.2² = 149.08 J