contestada

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what is the x position of the center of mass? m 2) what is the y position of the center of mass? m 3) what is the speed of the center of mass? m/s 4) when a fifth mass is placed at the origin, what happens to the horizontal (x) location of the center of mass? it moves to the right. it moves to the left. it does not move. it can not be determined unless you know the mass. 5) when a fifth mass is placed at the center of mass, what happens to the vertical (y) location of the center of mass? it moves up. it moves down. it does not move. it can not be determined unless you know the mass.

Respuesta :

Demiurgos
I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
[tex]R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i[/tex]
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
[tex]R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i[/tex]
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
[tex]R_x=0.96m[/tex]
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
[tex]R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i[/tex]
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
[tex]R_y=-0.84m[/tex]
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
[tex]v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}[/tex]
[tex]v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}[/tex]
Total velocity is:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
Once we calculate velocities we get:
[tex]v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}[/tex]
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing
Ver imagen Demiurgos
Ver imagen Demiurgos
batolisis

The answers to your questions are as follows

  1. The x position of the center of mass =  0.77 m
  2. The y position of the center of mass = -1.008 m
  3. The speed of the center of mass = 1.314 m/s
  4. The horizontal ( x ) location of the center of mass : It moves to the left
  5. The vertical ( y ) location of the center of mass :  It does not move

Calculating the (x) position of the center of mass

Xm = ( m₁x₁ + m₂x₂ + m₃x₃ + m₄x₄ )  / ( m₁ + m₂ + m₃ + m₄ )

     = (( 7.9*(-2.5 ) + 9.1*(-3.7) + 8.3 * 4.6 + 7.5 * 5.4 )) / ( 7.9 + 9.1 + 8.3 + 7.5 )

     = 0.77 m

Calculating the y position of the center of mass

Ym = ( m₁y₁ + m₂y₂ + m₃y₃ + m₄y₄ )  / ( m₁ + m₂ + m₃ + m₄ )

      = -1.008 m

Calculating the speed of the center of mass

Vm = [tex]\sqrt{Vx^2 + Vy^2}[/tex]

     where : Vx =  m₁vx₁ + m₂vx₂ + m₃vx₃ + m₄vx₄ )  / ( m₁ + m₂ + m₃ + m₄ )

                   Vy =  m₁vy₁ + m₂vy₂ + m₃vy₃ + m₄vy₄ )  / ( m₁ + m₂ + m₃ + m₄ )

Vx = -1.29 m/s

Vy = 0.25 m/s

Hence Vm = 1.314 m/s

When a fifth mass is placed at the origin,The horizontal ( x ) location of the center of mass moves to the left

When a fifth mass is placed at the center of mass The vertical ( y ) location of the center of mass :  It does not move

Hence we can conclude that the answers to your questions are as listed above.

Learn more about center of mass : https://brainly.com/question/16835885

Attached below is the missing part of the question

Ver imagen batolisis

Otras preguntas