Surface made by balls thrown in different directions

[Pushoam]

Homework Statement

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

 

[Orodruin]

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.

 

[Delta2]

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.

 

[Pushoam]

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).

 

[Pushoam]

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.

 

[Delta2]

(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)..

 

[Pushoam]

Thanks

 

[TSny]

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

 

[Pushoam]

${\sin {\theta} }^2+ {\cos{\theta} }^2 = 1$
$z(t) = v \cos{ \theta } t - \frac 1 2 get^2$ … (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.

 

[Delta2]

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.