Determine the location of the center of mass of a "L" whose thin vertical and horizontal members have the same length L and the same mass M. Use the formal definition to find the x and y coordinates, and check your result by doing the calculation with respect to two different origins, one in the lower left corner at the intersection of the horizontal and vertical members and one at the top of the vertical member.

(a) Origin at the lower left

x = ?

y = ?

(b) Origin at the top of the vertical member

x = ?

y = ?

a)  x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2

Explanation:

The center of mass of a body is the point where all external forces are applied, it is defined by

      = 1 / M ∑   = 1 /M ∫ x dm

      = 1 / M ∑ = 1 / M ∫ y dm

where M is the total body mass

Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis

a) let's start with the reference zero at the left end of the horizontal rod

let's use the concept of linear density

    λ = M / L = dm / dl

since the rod is on the x axis

     dl = dx

    dm = λ dx

let's calculate

      x_{cm} = M ∫ x λ dx = λ / M ∫ x dx

      x_{cm} = λ / M x² / 2

we evaluate between the lower integration limits x = 0 and upper x = L

      x_{cm} = λ / M (L² / 2 - 0)

  we introduce the value of the density that is cosntnate

     x_{cm} = (M / L) L² / 2M

     x_{cm} = L / 2

We repeat the calculation for verilla verilla

     λ = M / L = dm / dy

     y_{cm} = 1 / M ∫ y λ dy

     y_{cm} = λ M y² / 2

      = M/L  1/M (L² - 0)

     y_{cm}= L/2

b) we repeat the calculation for the origin the reference system is top of the vertical rod

     horizontal rod

        x_{cm} = 1 / M ∫λ x dx = λ/M   x² / 2

we evaluate between the lower limits x = 0 and the upper limit x = -L

      x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M

      x_{cm} = L / 2

vertical rod

      y_{cm} = 1 / M ∫y dm

      y_{cm} = λ / M ∫y dy

      y_{cm} = λ / M y2 / 2

we evaluate between the integration limits x = 0 and higher x = -L

      y_{cm} = (M / L) 1 / M ((-L)²/2 -0)

      y_{cm} = L / 2

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.