A uniform soda can of mass 0.135 kg is 12.4 cm tall and filled with 1.25 kg of soda (figure below). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. (Initially the soda can is full.)(a) What is the height h of the com (center of mass) of the can and contents initially. cm (b) What is the height h of the com of the can after the can loses all the soda? cm (c) What happens to h as the soda drains out? ---Select--- rises to the top decreases then rises again stays the same decreases to the bottom (d) If x is the height of the remaining soda at any give instant, find x when the com reaches its lowest point. cm
To solve this problem, let's break it down step-by-step:
(a) The center of mass of the can and the contents initially is the weighted average of their respective center of masses. Since the soda is completely filling the can, its center of mass is at its midpoint. The height of the soda is given as h, so the center of mass of the soda is at h/2.
For the can, we can consider it as a uniform rod with its center of mass at its midpoint. The height of the can is given as 12.4 cm, so the center of mass of the can is at 12.4/2 = 6.2 cm.
To find the center of mass of the can and contents initially, we can use the mass-weighted average formula:
h_com_initial = (m_soda * (h/2) + m_can * (6.2 cm)) / (m_soda + m_can)
where m_soda is the mass of the soda (1.25 kg) and m_can is the mass of the can (0.135 kg).
Substituting the values into the formula:
h_com_initial = (1.25 kg * (h/2) + 0.135 kg * (6.2 cm)) / (1.25 kg + 0.135 kg)
Simplifying the equation gives us:
h_com_initial = (0.625h + 0.837) / 1.385
(b) After the can loses all the soda, it becomes empty and the center of mass of the can will be at its own midpoint. So the height of the center of mass of the can after losing all the soda is equal to half of its height:
h_com_final = 6.2 cm / 2
(c) As the soda drains out, the height of the center of mass, h_com, decreases. This happens because the soda was initially at a higher position, contributing to the higher center of mass. As it drains out, the remaining mass is concentrated lower in the can, causing the center of mass to lower as well.
(d) To find the height, x, of the remaining soda when the center of mass reaches its lowest point, we need to find the point where the center of mass is at its lowest position.
To do this, we set the derivative of h_com with respect to x equal to zero and solve for x:
d(h_com)/dx = 0
By differentiating the formula for h_com, we have:
d(h_com)/dx = (m_can * 6.2 cm - m_soda * (h - x)) / (m_soda + m_can)^2
Setting this expression equal to zero and solving for x, we have:
m_can * 6.2 cm - m_soda * (h - x) = 0
Solving for x:
x = (m_can * 6.2 cm) / m_soda + h
Substituting the given values, x is equal to:
x = (0.135 kg * 6.2 cm) / 1.25 kg + h
Simplifying the equation gives us:
x = 0.675 cm + h
To solve this problem, we need to consider the center of mass (COM) of both the can and the soda separately.
(a) To find the initial COM of the can and contents, we can use the principle of equilibrium. The COM is the point where the system can be balanced. Assuming the can's mass is evenly distributed, the COM of the can will be at its center, which is half of its total height. So, the COM of the can is 12.4 cm/2 = 6.2 cm.
For the soda, since it is filled uniformly, its COM will also be at its center. Therefore, the COM of the soda is also 12.4 cm/2 = 6.2 cm.
To find the COM of the can and contents together, we can apply the principle of weighted average. The mass of the can is 0.135 kg and the mass of the soda is 1.25 kg. Since the COM of the can and the soda are at the same height, their individual masses can be used as weight factors in the average formula.
COM_initial = (m_can * h_can + m_soda * h_soda) / (m_can + m_soda)
Given:
m_can = 0.135 kg
m_soda = 1.25 kg
h_can = 6.2 cm
h_soda = 6.2 cm
COM_initial = (0.135 kg * 6.2 cm + 1.25 kg * 6.2 cm) / (0.135 kg + 1.25 kg)
COM_initial = 1.2355 kg.cm / 1.385 kg
COM_initial ≈ 0.891 cm
Therefore, the height of the COM of the can and contents initially is approximately 0.891 cm.
(b) After all the soda drains out, the only mass remaining is that of the can, and its COM will stay at the same height. Therefore, the height of the COM of the can after the soda drains out remains at 6.2 cm.
(c) As the soda drains out, the height of the COM will decrease. This happens because the can's mass is concentrated at the bottom, while the soda mass is distributed higher up, which pulls the COM downward. So, the correct answer is "decreases then rises again."
(d) To find the height (x) of the remaining soda when the COM reaches its lowest point, we can set up a balance equation between the can and the remaining soda.
m_can * h_can = m_soda * (h_can + x)
Given:
m_can = 0.135 kg
h_can = 6.2 cm
We need to solve for x.
0.135 kg * 6.2 cm = 1.25 kg * (6.2 cm + x)
0.837 kg.cm = 1.25 kg * (6.2 cm + x)
0.837 kg.cm = 7.75 kg.cm + 1.25 kg * x
0.837 kg.cm - 7.75 kg.cm = 1.25 kg * x
-6.913 kg.cm = 1.25 kg * x
x = (-6.913 kg.cm) / (1.25 kg)
x = -5.53 cm
Therefore, when the COM reaches its lowest point, the height of the remaining soda (x) is approximately -5.53 cm. Note that the negative sign indicates that the remaining soda is located below the original COM position.
a) The center of mass of the can is halfway up; the center of mass of the soda in the can is also halfway up when the can is full of soda, so the center of mass is
(0.135*(12.4/2) + 1.25*(12.4/2))/(0.135+1.25)
b) After the can loses all the soda, the center of mass is the center of mass of the can, which is halfway up or 12.4/2
c)
x/12.4 is the percentage of soda that has been drained out.
The center of mass of the remaining soda is x/2. The mass of the remaining soda is 1.25*(x/12.4)
Therefore, the COM of the system of the can + soda is:
h = (0.135*(12.4/2) + 1.25*(x/12.4)*(x/2))/(0.135+1.25*(x/12.4)
Graph this function, and see what happens to it as a function of h
d) Evaluate dh/dx = 0