Center of Mass Projectile Problem

In summary, the projectile has a center of mass that is at the origin and it falls vertically when the y coordinate is 0. The x coordinate of the center of mass is at 6.93 seconds, which is the position of fragment 1.
  • #1
itsme24
8
0
Hi, sorry stuck again!

Here is the problem:

A projectile of mass 19.4 kg is fired at an angle of 57.0 degrees above the horizontal and with a speed of 81.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance.

And here is what I did:

I first found the center of mass when y = 0 since the center of mass follows the trajectory of the parabola I used the equation:

y(x) = tan(theta)x - .5g(x/(V_i*cos(theta))^2

y(x) = 0
theta = 57 degrees
g= 9.8m/s^2
V_i = 81.0 m/s so...

0 = tan(57)x - 4.9(x/(8.1*cos(57))^2
0 = 1.54x - 0.00252x^2

quadratic:
x = 0, 776.1m = center of mass

Then it says that the first fragment drops vertically at the highest point and this is partly where I get confused since with my math I assumed that means straight down, correct me if I'm wrong. So I found t when V_y = 0 to find x at that point.

V_y = 0 = V_yi + a_y*t

0= 81.0*sin(57) - 9.8t
t= 6.93s

x when t = 6.93s which should give me the position fragment 1 landed:

x_f1 = x_i + V_ix*t + 0.5a_x*t^2
x_f1 = 0 + 81.0cos(57) + 0
x_f1 = 305.72m

Then I just had to find x of fragment 2:

x_cm = (m_f1*x_f1) + (m_f2*x_f2) / (m_f1 + m_f2)

so, I made the center of mass the 0 coordinate:

776.1m = [(9.7kg*-470.4m) + (9.7kg*x_f2)] / (9.7kg*2)

x_f2 = 2020m, which turned out to be wrong! :eek:

I'm sorry this problem is so long :frown:
 
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  • #2
Why did [tex]x_{f1}[/tex] become -470.4m, when it clearly states it's 305.27m above?

Also [tex]m_1 + m_2[/tex] is 19.4, not 9.7
 
  • #3
ya sorry I had it written 9.7kg*2, I did use 19.4.

But for the x_f1 I put it to -470.4m because I put the center of mass that I found as the origin in a new axes for calculating x_f2's position.

so 776.1m became 0, which means that 776.1m-305.72m became x_f1 and since it's to the left of the new origin it would be negative. Should I not have done that?
 
  • #4
If I left it at 305.27m then the math would be

776.1m = [(9.7kg*305.27m) + 9.7kg*x_f2] / 19.4kg = 1246.93 m and that answer was not correct either :(
 

Related to Center of Mass Projectile Problem

What is a Center of Mass Projectile Problem?

A Center of Mass Projectile Problem is a type of physics problem that involves determining the trajectory of an object in motion, taking into account the object's center of mass. This problem is commonly encountered in fields such as mechanics and engineering.

How do you calculate the center of mass for a projectile?

The center of mass for a projectile can be calculated by finding the weighted average of the positions of all the individual particles that make up the projectile. This can be done using the formula xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where x is the position and m is the mass of each particle.

Why is the center of mass important in projectile motion?

The center of mass is important in projectile motion because it is the point at which all the mass of the object can be considered to be concentrated. This point behaves as if all the mass of the object is located at this single point, making it easier to analyze the motion of the object.

How does the center of mass affect the trajectory of a projectile?

The center of mass affects the trajectory of a projectile by determining the point around which the object rotates and the direction in which it moves. The center of mass also affects the stability of the projectile, as a higher or lower center of mass can cause the projectile to tip or flip during flight.

Can the center of mass change during projectile motion?

Yes, the center of mass can change during projectile motion if there is a change in the distribution of mass within the object or if external forces act on the object. For example, a rocket's center of mass will shift as fuel is burned and ejected during flight.

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