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2. (8pts) A system of balls connected by massless rods is initially at rest on a small table as
shown in the figure below. All balls have a mass of m = 2.0 kg except for the ball at (0.2m, 0)
which has a mass of M = 7.0 kg. a. Find the x-, y-coordinates of the center of mass (center of
gravity) of the system. Give your answers to two significant figures. b. Is the system in a stable
or unstable equilibrium? Why? c. Is the system in static equilibrium about an axis through the
ball at (0.2m, 0)? Why?
Solution
a)
Xcm=[(m1*x1)+(m2*x2)+(m3*x3)+(m4*x4)+(m5*x5)+(m6*x6)+(m7*x7)]/(m1+m2+m3+m5+
m6+m7)
Xcm=[(2*0.1)+(2*0.1)+(2*0.15)+(2*0.25)+(2*0.3)+(2*0.35)+(7*0.2)]/(2+2+2+2+2+2+7)
=(0.2+0.2+0.3+0.5+0.6+0.7+1.4)/(12+7)
=0.21
Yc.m=[(m1*y1)+(m2*y2)+(m3*y3)+(m4*y4)+(m5*y5)+(m6*y6)+(m7*y7)]/(m1+m2+m3+m5+
m6+m7)
=(2*0)+(2*0.1)+(2*0.2)+(2*0.1)+(2*0)+(2*0.2)+(7*0)]/(2+2+2+2+2+2+7)
=(0+0.2+0.4+0.2+0+0.4+0)/(12+7)
=1.2/19
=0.06
center of mass (Xcm, Ycm)=(0.21, 0.06)
b)
stable equilibrium, because initially masses are at rest
c)
the system is not in a
because for static equilibrium
Fnet=0
Tnet=0
but here net torque may not be zero

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2024-NATIONAL-LEARNING-CAMP-AND-OTHER.pptx
 

2. (8pts) A system of balls connected by massless rods is initially a.pdf

  • 1. 2. (8pts) A system of balls connected by massless rods is initially at rest on a small table as shown in the figure below. All balls have a mass of m = 2.0 kg except for the ball at (0.2m, 0) which has a mass of M = 7.0 kg. a. Find the x-, y-coordinates of the center of mass (center of gravity) of the system. Give your answers to two significant figures. b. Is the system in a stable or unstable equilibrium? Why? c. Is the system in static equilibrium about an axis through the ball at (0.2m, 0)? Why? Solution a) Xcm=[(m1*x1)+(m2*x2)+(m3*x3)+(m4*x4)+(m5*x5)+(m6*x6)+(m7*x7)]/(m1+m2+m3+m5+ m6+m7) Xcm=[(2*0.1)+(2*0.1)+(2*0.15)+(2*0.25)+(2*0.3)+(2*0.35)+(7*0.2)]/(2+2+2+2+2+2+7) =(0.2+0.2+0.3+0.5+0.6+0.7+1.4)/(12+7) =0.21 Yc.m=[(m1*y1)+(m2*y2)+(m3*y3)+(m4*y4)+(m5*y5)+(m6*y6)+(m7*y7)]/(m1+m2+m3+m5+ m6+m7) =(2*0)+(2*0.1)+(2*0.2)+(2*0.1)+(2*0)+(2*0.2)+(7*0)]/(2+2+2+2+2+2+7) =(0+0.2+0.4+0.2+0+0.4+0)/(12+7) =1.2/19 =0.06 center of mass (Xcm, Ycm)=(0.21, 0.06) b) stable equilibrium, because initially masses are at rest c) the system is not in a because for static equilibrium
  • 2. Fnet=0 Tnet=0 but here net torque may not be zero