1. The figure below shows a cubical box that has been constructed from uniform m
ID: 1417692 • Letter: 1
Question
1. The figure below shows a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length L =50 cm. Find the coordinates of the center of mass of the box.
cm (x-coordinate)
cm (y-coordinate)
cm (z-coordinate)
2. A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 150 ms.
(a) How far below the release point is the center of mass of the two stones at t = 350 ms? (Neither stone has yet reached the ground.)
m
(b) How fast is the center of mass of the two-stone system moving at that time?
m/s
Explanation / Answer
Solution:
The top is open. So we ned to find the x, y , z coordinates of the other 5 faces.
L=50 cm .
For side in the YZ plane : (0, 25, 25)
For side in the XZ plane , (25 , 0 , 25)
For the side in the XY plane , ( 25 , 25 , 0 )
For the side in the for the front face : (50 , 25 , 25 )
For the right face : ( 25 , 50 , 25 )
X cm = ( m1x1+m2x2 + 3x3 + m4 x4 + m5 x5) / (m1+m2+m3+m4 +m5)
= ( m(0) + m(25) +m(25)+m(50) + m(25) / (m+m+m+m+m)
= 125 m / 5m = 25
Y cm = m(25) + m(0) + m(25)+m(25) +(50) / (m+m+m+m+m) = 125 m /5m = 25
Zcm = m(25)+m(25) +m(0) +m(25)+m(25) / (m+m+m+m+m) =100 m/5m = 20
2) Y = 1/gt^2
For the 1st stone of mass m, time = t1 = 350ms = 0.35 s
y1 = 1/2 gt1^2 = 1/2 * 9.8 (0.35)^2 = 0.6m
for the 2nd stone , of mass twice the first, time = t2 = 0.35 - 0.15 =0.2 s
y2 = 1/2gt2^2 = 1/2 * 9.8 * (0.2)^2 = 0.196 m
The center of mass = Y cm = m1y1+m2y2/(m1+m2)
= m(0..6) + 2(m)(0.196) / (m+3m) = 0.33 m
b) v1 = gt1 = 9.8 * 0..35 = 3.43 m/s
v2 = gt2 = 9.8 * (0.35-0.15) = 1.96 m/s
Vcm = m1v1+m2v2 /(m1+m2) = (m)(3.43) + 2m(1.96) / (m+2m) = 2.45 m/s
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