Surface made by balls thrown in different directions

In summary, the problem involves throwing balls in different directions from a fixed point and determining the shape and motion of the balls. The equations for the motion of the balls in cylindrical coordinates are given, and the correct answer is (c) - a spherical surface with radius r = vt and center going downwards with acceleration g. This can be proven by considering a frame that accelerates downwards with acceleration g and showing that the position vector of the balls is independent of the angle and increases with time. The equation of the sphere in vector form is also provided.
  • #1
Pushoam
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Homework Statement



upload_2017-12-22_12-23-28.png

Homework Equations


I consider four directions from the point of throwing the balls, east, west, up, down. At any given time, the balls will be at different distance from this point, (the upward ball will travel less distance than the downward ball). So, it could not be a sphere with fixed center.

I think I should solve it instead of guessing the answer.

Taking cylinderical coordinate system taking origin at the point of throw , in which z- axis is the axis is the the axis along which gravitational force acts.

## z(t) = h + v_zt - \frac 1 2 gt^2 ## … (1) ## v_z = v \cos{ \theta } ## ……(2), where ##\theta## is the angle ##\vec v## makes with z-axis.

## z(t) = v \cos{ \theta }~ t - \frac 1 2 g t^2 ## … (1)

## s(t) = v \sin{ \theta }~t ## ……..(3)If there had been no gravitational force, the answer would have been a sphere centred at the point of throw with increasing radius i.e. opinion (a).

Because of the gravitational force, I guess that the center should go down and hence option (c).

But I don’t know how to show it using the above equations.

The Attempt at a Solution

 

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  • #2
Hint: I suggest looking at the problem in a frame that accelerates downwards with acceleration ##g## and then figure out how this relates to the lab frame.

Edit: To be clear, your answer (c) is correct.
 
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  • #3
I also think (c) is the correct answer.

To be honest i am not sure in what frame and which coordinate system (cylindrical or Cartesian probably) you should chose but the line of thinking would be to express the position vector for a ball that is thrown at angle ##\theta## which will be ##\vec{R_\theta(t)}## in that frame and coordinate system, and then prove that its magnitude ##|\vec{R_\theta}(t)|## is independent of the angle ##\theta## and increasing function of time t. Probably in the expression for the magnitude will appear some ##\cos^2\theta, \sin^2\theta## terms which will cancel out.
 
  • #4
Orodruin said:
Hint: I suggest looking at the problem in a frame that accelerates downwards with acceleration ggg and then figure out how this relates to the lab frame.
Trying to visualize it this way.

When there is no acceleration, the surface will be spherical with radius r = vt.

Considering a frame which is going upwards with speed u. This fellow will see that the spherical surface is coming down with speed u.

Considering a frame which is going upwards with acceleration g. This fellow will see that the spherical surface is coming down with acceleration g.So, the answer is : the sphere surface with radius r = vt with center going downwards with acceleration g. Hence, the correct option is (c).
 
  • #5
Delta² said:
To be honest i am not sure in what frame and which coordinate system (cylindrical or Cartesian probably) you should chose but the line of thinking would be to express the position vector for a ball that is thrown at angle θ which will be ##\vec{R_\theta(t)}## in that frame and coordinate system, and then prove that its magnitude##|\vec{R_\theta}(t)| ##is independent of the angle θ and increasing function of time t. Probably in the expression for the magnitude will appear some ##\cos^2\theta, \sin^2\theta## terms which will cancel out.

To solve analytically,

With respect to a co - ordinate frame S' moving downwards with acceleration magnitude g and whose origin coincides with the point of throw at the time of throw. In this frame, the balls are not accelerating.

## \vec r'_{\theta }(t) = \vec v' t ## ...(1) where ## \vec v' ## = initial velocity of throw i.e.##\vec v##.

With respect to a frame S, which is fixed and coincides with the point of throw,

## \vec r(t) = \vec r'(t) + \vec {OO'}(t) ## ...(2)

Where ## \vec{OO'}(t) ## is position vector of the origin of S' frame i.e. O' from the origin of S frame i.e. O.

Now, ## \vec {OO'}(t) = \frac 1 2 \vec g t^2 ## ...(3)

So, using (1), (2),(3)

## \vec r(t) = \vec v t + \frac 1 2 \vec g t^2 ## ...(4)

But, I don’t know how to convert (4) to " the surface is spherical with radius r = vt and center going downwards with acceleration magnitude g? Please help me here.
 
  • #6
(4) if you write it as ##\vec{r_\theta(t)}-\vec{R_0(t)}=\vec{v_\theta}t## ,w here ##\vec{R_0(t)}=\frac{1}{2}\vec{g}t^2## denotes a sphere with radius ##|\vec{r_\theta(t)}-\vec{R_0(t)}|=|\vec{v_\theta}t|=vt## (since all ##v_\theta## have equal magnitude ##v##, that is same initial speed) and center at the point of the vector ##\vec{R_0(t)}##.

(Note: Using vectors the equation of a sphere with center at the point of vector ##\vec{R_0}## is

##\vec{R}-\vec{R_0}=\vec{C}## where C is a vector which has constant magnitude that is ##|\vec{C}|## is independent of the spatial variables x,y,z and ##\vec{R}=x\hat i+y\hat j+z\hat k## is the position vector (expressed in Cartesian coordinates in this case)..
 
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  • #7
Thanks
 
  • #8
Pushoam said:
## z(t) = v \cos{ \theta }~ t - \frac 1 2 g t^2 ## … (1)

## s(t) = v \sin{ \theta }~t ## ……..(3)

But I don’t know how to show it using the above equations.
Orodruin's and Delta2's solutions are very nice. If you want to show it with equations (1) and (3), use these equations to set up
cos2θ + sin2θ = 1

(and you have rotational symmetry about the z axis)
 
  • #9
## {\sin {\theta} }^2+ {\cos{\theta} }^2 = 1 ##
## z(t) = v \cos{ \theta } t - \frac 1 2 get^2 ## … (1) ##

## s(t) = v \sin{ \theta }t ## ……..(3) ##

## v^2 t^2 = \{z(t) + \frac 1 2 get^2\} ^2+ \{ s(t)\}^2 ## ...(4)

The above is the equation of sphere whose center goes down with acceleration - magnitude g.

Thanks.
 
  • #10
Just to say some things to clarify more my post, ##v_\theta## is the velocity of the sphere that is thrown making angle ##\theta## with the horizontal axis (and NOT the ##\theta## component of velocity in a polar or cylindrical or spherical coordinate system) and it should be ##v_{\theta,\phi}## if we are working in 3 dimensions as the problem implies.

And also to be more accurate regarding the equation of sphere in vector notation, in order for the endpoints of a vector ##\vec{R}=f(x,y,z)\hat i+g(x,y,z)\hat j+ q(x,y,z)\hat k## to form a sphere with center the endpoint of a vector ##\vec{R_0}## the condition that must hold is ##|\vec{R}-\vec{R_0}|=c## where c is a constant number independent of the variables x,y,z or more general independent of the spatial coordinates of the coordinate system that vector R is expressed.
 
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1. What is a "surface made by balls thrown in different directions"?

A "surface made by balls thrown in different directions" refers to a mathematical concept known as a paraboloid. It is a curved surface that is created by rotating a parabola around its axis. In this case, the parabola represents the path of a ball thrown in a certain direction.

2. How is the surface made by balls thrown in different directions useful in science?

The surface made by balls thrown in different directions has many practical applications in science. For example, it can be used to model the trajectory of a projectile in physics, the shape of a satellite dish in engineering, and the shape of a bowl in architecture.

3. How do the directions and speeds of the balls affect the resulting surface?

The directions and speeds of the balls have a significant impact on the shape of the resulting surface. The angle at which the balls are thrown will determine the steepness of the parabola, while the speed will affect the overall size and shape of the surface.

4. Are there any real-life examples of surfaces made by balls thrown in different directions?

Yes, there are many real-life examples of surfaces made by balls thrown in different directions. Some examples include the shape of a fountain, the shape of a waterfall, and the shape of a splash in a pool. These surfaces are all created by the motion of objects moving in different directions and speeds.

5. What other factors can influence the shape of the surface?

Aside from the directions and speeds of the balls, other factors that can influence the shape of the surface include air resistance, gravity, and the surface on which the balls are thrown. These factors can affect the trajectory of the balls and ultimately impact the resulting shape of the surface.

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