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center of mass 1. find the center of mass (relative to (0,0)) for the s…

Question

center of mass

  1. find the center of mass (relative to (0,0)) for the spheres with the following masses and locations: m1 = 5kg, (1,1) m2 = 10 kg, (3,1) m3 = 15kg, (1,6).
  2. find the center of mass of the following particles (drawn large so they can be seen): image of particles with masses and distances
  3. an old go - kart with a mass of 300 kg is traveling in a straight line at 80 m/s. it is followed by a 4 - wheeler with mass of 200 kg moving at 60 m/s. how fast is the center of mass moving?
  4. a 1500 kg vw is heading 40 m/s in a straight line. a 4000 kg cadillac is heading directly for it at 60 m/s. find the velocity (magnitude and direction) of the center of mass.
  5. a 1500 kg car is at rest. at the instant it starts to move (with an acceleration of 3.5m/s²), a truck (m = 3000kg) traveling at a constant speed of 12 m/s passes it. at t = 3 seconds:

a. how far is the center of mass of the vehicles, relative to the starting point of the car?
b. what is the speed of the center of mass of the vehicles?

  1. a rock, of mass m, is dropped at t = 0 seconds. two seconds later a stone, of mass 2m, is dropped. at t = 3 seconds (assume neither hits the ground):

a. what is the center of mass of the rock and stone relative to the drop point?
b. how fast is the center of mass going at this time?

  1. calculate the vcm before the collision and then calculate the vcm after the collision. (show all work for this problem)

before collision: image of two masses with initial and final velocities for one mass
after collision

Explanation:

Step1: Recall center - of - mass formula in 2D

The formula for the $x$ - coordinate of the center of mass $\bar{x}=\frac{\sum_{i = 1}^{n}m_ix_i}{\sum_{i = 1}^{n}m_i}$ and for the $y$ - coordinate $\bar{y}=\frac{\sum_{i = 1}^{n}m_iy_i}{\sum_{i = 1}^{n}m_i}$

Step2: Calculate for problem 1

Given $m_1 = 5kg$, $x_1=1$, $y_1 = 1$, $m_2=10kg$, $x_2 = 3$, $y_2=1$, $m_3 = 15kg$, $x_3=1$, $y_3 = 6$
First, calculate the total mass $M=m_1 + m_2+m_3=5 + 10+15=30kg$
For the $x$ - coordinate of the center of mass:
$\bar{x}=\frac{m_1x_1+m_2x_2+m_3x_3}{M}=\frac{5\times1 + 10\times3+15\times1}{30}=\frac{5 + 30+15}{30}=\frac{50}{30}=\frac{5}{3}$
For the $y$ - coordinate of the center of mass:
$\bar{y}=\frac{m_1y_1+m_2y_2+m_3y_3}{M}=\frac{5\times1+10\times1 + 15\times6}{30}=\frac{5+10 + 90}{30}=\frac{105}{30}=\frac{7}{2}$

Step3: Calculate for problem 2

Assume the origin at the center of the 5 - kg particle.
$m_1 = 20kg$, $x_1=-6m$, $y_1 = 0$, $m_2=5kg$, $x_2 = 0$, $y_2=0$, $m_3 = 10kg$, $x_3=2m$, $y_3=-3m$
Total mass $M=m_1 + m_2+m_3=20 + 5+10=35kg$
$x$ - coordinate of the center of mass:
$\bar{x}=\frac{m_1x_1+m_2x_2+m_3x_3}{M}=\frac{20\times(-6)+5\times0 + 10\times2}{35}=\frac{-120 + 20}{35}=\frac{-100}{35}=-\frac{20}{7}m$
$y$ - coordinate of the center of mass:
$\bar{y}=\frac{m_1y_1+m_2y_2+m_3y_3}{M}=\frac{20\times0+5\times0+10\times(-3)}{35}=\frac{-30}{35}=-\frac{6}{7}m$

Step4: Calculate for problem 3

The formula for the velocity of the center of mass of a two - particle system is $v_{cm}=\frac{m_1v_1+m_2v_2}{m_1 + m_2}$
Given $m_1 = 300kg$, $v_1 = 80m/s$, $m_2=200kg$, $v_2 = 60m/s$
$v_{cm}=\frac{300\times80+200\times60}{300 + 200}=\frac{24000+12000}{500}=\frac{36000}{500}=72m/s$

Step5: Calculate for problem 4

Let the direction of the VW be positive. Then $m_1 = 1500kg$, $v_1 = 40m/s$, $m_2=4000kg$, $v_2=-60m/s$
$v_{cm}=\frac{m_1v_1+m_2v_2}{m_1 + m_2}=\frac{1500\times40+4000\times(-60)}{1500 + 4000}=\frac{60000-240000}{5500}=\frac{-180000}{5500}=-\frac{360}{11}\approx - 32.73m/s$
The magnitude is $\frac{360}{11}m/s\approx32.73m/s$ and the direction is the same as the direction of the Cadillac.

Step6: Calculate for problem 5a

For the car: Using $x_1=\frac{1}{2}at^2$, with $a = 3.5m/s^2$ and $t = 3s$, $x_1=\frac{1}{2}\times3.5\times3^2=\frac{1}{2}\times3.5\times9 = 15.75m$
For the truck: Using $x_2=v_2t$, with $v_2 = 12m/s$ and $t = 3s$, $x_2=12\times3 = 36m$
$m_1 = 1500kg$, $m_2=3000kg$
Total mass $M=m_1 + m_2=1500+3000 = 4500kg$
$x_{cm}=\frac{m_1x_1+m_2x_2}{M}=\frac{1500\times15.75+3000\times36}{4500}=\frac{23625+108000}{4500}=\frac{131625}{4500}=29.25m$

Step7: Calculate for problem 5b

For the car: $v_1=at=3.5\times3 = 10.5m/s$
For the truck: $v_2 = 12m/s$
$v_{cm}=\frac{m_1v_1+m_2v_2}{M}=\frac{1500\times10.5+3000\times12}{4500}=\frac{15750+36000}{4500}=\frac{51750}{4500}=11.5m/s$

Step8: Calculate for problem 6a

For the rock: $h_1=\frac{1}{2}gt^2$, with $t = 3s$, $h_1=\frac{1}{2}g\times3^2=\frac{9}{2}g$
For the stone: $t'=3 - 2=1s$, $h_2=\frac{1}{2}g\times1^2=\frac{1}{2}g$
$m_1 = M$, $m_2=2M$
Total mass $M_{total}=M + 2M=3M$
$h_{cm}=\frac{m_1h_1+m_2h_2}{M_{total}}=\frac{M\times\frac{9}{2}g+2M\times\frac{1}{2}g}{3M}=\frac{\frac{9}{2}Mg+Mg}{3M}=\frac{\frac{9 + 2}{2}Mg}{3M}=\frac{11}{6}g\approx18.17m$ (taking $g = 10m/s^2$)

Step9: Calculate for problem 6b

For the rock: $v_1=gt=3g$
For the stone: $v_2=g\times1 = g$
$v_{cm}=\frac{m_1v_1+m_2v_2}{M_{total}}=\frac{M\times3g+2M\times g}{3M}=\frac{3Mg+2Mg}{3M}=\frac{5}{3}g\approx16.67m/s$

Step10: Calculate for problem 7 (before collision)

$m_1 = 4kg$, $…

Answer:

  1. $x=\frac{5}{3}$, $y=\frac{7}{2}$
  2. $x =-\frac{20}{7}m$, $y=-\frac{6}{7}m$
  3. $72m/s$
  4. Magnitude: $\frac{360}{11}m/s\approx32.73m/s$, Direction: same as the Cadillac

5a. $29.25m$
5b. $11.5m/s$
6a. $\frac{11}{6}g\approx18.17m$
6b. $\frac{5}{3}g\approx16.67m/s$

  1. Before collision: $\frac{10}{7}m/s$, After collision: $\frac{10}{7}m/s$