物理英文題目求解答?
2015-11-09 9:58 am
A metal soda can of uniform composition has a mass of 0.110 kg and is 14.7 cm tall The can is filled with 1.04 kg of soda. Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height h of the center of mass of the can and contents (a) initially and (b) after the can loses all the soda? (c) If x is the height of the remaining soda at any given instant, find x when the center of mass reaches its lowest point.
回答 (1)
2015-11-09 9:42 pm
L = 14.7 cm
m1 = 0.11 kg
m2 = 1.04*x/14.7 = 0.07075*x = k*x
(a) initially
Ybar(x) = (m1*L/2 + m2*x/2)/(m1+m2)
= (m1*L + kx^2)/2(m1 + kx)
Ybar(L) = (m1+kL)L/2(m1+kL) = L/2 = 7.35 cm
(b) after the can loses all the soda?
Ybar(0) = m1*L/2m1 = L/2 = 7.35 cm
(c) If x is the height of the remaining soda at any given instant, find
x when the center of mass reaches its lowest point.
2*Ybar*(m1+kx) = m1*L + kx^2
Set Ybar' = 0:
2Ybar'*(m1+kx) + 2k*Ybar = 2kx
2k*Ybar = 2kx
x = Ybar = (m1*L + kx^2)/2(m1 + kx)
m1*L + kx^2 = 2m1*x + 2kx^2
0 = kx^2 + 2m1*x - m1*L
x = {-m1 + √(m1^2 + m1*L*k)}/k
= {-0.11 + √(0.11^2 + 0.11*14.7*0.07075)}/0.07075
= 3.47 cm
m1 = 0.11 kg
m2 = 1.04*x/14.7 = 0.07075*x = k*x
(a) initially
Ybar(x) = (m1*L/2 + m2*x/2)/(m1+m2)
= (m1*L + kx^2)/2(m1 + kx)
Ybar(L) = (m1+kL)L/2(m1+kL) = L/2 = 7.35 cm
(b) after the can loses all the soda?
Ybar(0) = m1*L/2m1 = L/2 = 7.35 cm
(c) If x is the height of the remaining soda at any given instant, find
x when the center of mass reaches its lowest point.
2*Ybar*(m1+kx) = m1*L + kx^2
Set Ybar' = 0:
2Ybar'*(m1+kx) + 2k*Ybar = 2kx
2k*Ybar = 2kx
x = Ybar = (m1*L + kx^2)/2(m1 + kx)
m1*L + kx^2 = 2m1*x + 2kx^2
0 = kx^2 + 2m1*x - m1*L
x = {-m1 + √(m1^2 + m1*L*k)}/k
= {-0.11 + √(0.11^2 + 0.11*14.7*0.07075)}/0.07075
= 3.47 cm
收錄日期: 2021-04-30 20:07:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151109015806AAlzT3D