Simple center of mass problem but i cant figure it out

In summary, the center of mass of a uniform, L-shaped iron rod with sides X = 0.8 m and Y = 0.5 m, respectively, located at (x,y) = (0,0), can be found by taking the corner as the origin and finding the coordinates (x,y) = (0.2667m, 0.1667m). The C.O.M of the horizontal leg is located at (0.4,0) and the C.O.M of the vertical leg is located at (0,0.25). The C.M coordinates are determined by dividing the mass of the horizontal leg (m_{h} = 0.8\rho) and the mass of
  • #1
huk
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I feel silly for ask you for help on this simple center of mass problem but i got wrong answers...

Where is the center of mass of a uniform, L-shaped iron rod of sides X = 0.8 m and Y = 0.5 m, respectively? Take the corner to be at (x,y) = (0,0), with the X and Y sides along those axes, respectively. (Assume that the rod is so narrow that the dimensions of the outer bend are the same as those of the inner bend of the L.)

i thought the answer is x = .2667m and y = .1667m perhaps my answers are wrong?
 
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  • #2
Now, the C.O.M of the horizontal leg is (0.4,0), for the vertical leg, (0,0.25)
(Agreed?)
Let [tex]m_{h}=0.8\rho[/tex] be the mass of the horizontal leg, and [tex]m_{v}=0.5\rho[/tex] the mass of the vertical leg ([tex]\rho[/tex] being the density).
Hence, the coordinates of the C.M of the L-shaped object fulfill:
[tex](x_{C.M},y_{C.M})=\frac{m_{h}(0.4,0)+m_{v}(0,0.25)}{m_{h}+m_{v}}[/tex]
 
  • #3


Don't feel silly for asking for help on a simple center of mass problem. Sometimes, even the simplest problems can be tricky to solve. It's always better to ask for help and get the correct answer, rather than struggling on your own and potentially getting the wrong answer.

Regarding your specific problem, your answers for the center of mass coordinates seem to be incorrect. To find the center of mass of a uniform object, we can use the formula:

x_cm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn)

Where x_cm is the x-coordinate of the center of mass, m is the mass of each part of the object, and x is the x-coordinate of each part.

For this L-shaped iron rod, we can divide it into two parts - the longer side with length 0.8 m and the shorter side with length 0.5 m. The longer side has a mass of 0.8 m, and its x-coordinate is 0.4 m (half of its length). The shorter side has a mass of 0.5 m, and its x-coordinate is also 0.4 m (half of its length).

Plugging these values into the formula, we get:

x_cm = (0.8*0.4 + 0.5*0.4) / (0.8 + 0.5) = 0.36 m

This means that the center of mass of the L-shaped rod is located at x = 0.36 m. Similarly, we can find the y-coordinate of the center of mass using the same formula.

y_cm = (m1y1 + m2y2 + ... + mny) / (m1 + m2 + ... + mn)

For this L-shaped rod, the y-coordinate of the longer side is 0.4 m (half of its length), and the y-coordinate of the shorter side is 0 m (since it is along the x-axis). Plugging these values into the formula, we get:

y_cm = (0.8*0.4 + 0.5*0) / (0.8 + 0.5) = 0.256 m

This means that the center of mass of the L-shaped rod is located at y = 0.256 m.

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FAQ: Simple center of mass problem but i cant figure it out

What is the center of mass and why is it important in physics?

The center of mass is the point at which an object's mass can be considered to be concentrated. It is important in physics because it helps determine an object's stability and how it will behave when subjected to external forces.

How do you calculate the center of mass for a simple object?

To calculate the center of mass for a simple object, you need to find the average position of all the individual particles that make up the object. This can be done by taking into account the mass and distance of each particle from a reference point.

Can you explain the concept of rotational equilibrium in relation to the center of mass?

Rotational equilibrium is a state in which an object is not experiencing any rotation. The center of mass plays a crucial role in determining whether an object is in rotational equilibrium or not. If the forces acting on an object are balanced and passing through the center of mass, the object will be in rotational equilibrium.

How does the distribution of mass affect the location of the center of mass?

The distribution of mass affects the location of the center of mass by shifting it towards the heavier side of the object. This means that the center of mass will be closer to the side with more mass, especially if the mass is concentrated in a particular area.

What are some real-life applications of the concept of center of mass?

The concept of center of mass is used in many real-life applications such as designing stable structures, determining the balance of vehicles, and analyzing the movements of athletes in sports. It is also used in astronomy to study the motion of celestial bodies and in engineering to design structures that can withstand external forces.

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