Homework help (normal and tangential coordinate problem)

[nineeyes]

Problem : A baseball player releases a ball with v_0= 100 ft/sec at an angle of \theta= 30 degrees. Determine the radius of curvature of the trajectory (a) just after release and (b) at the apex. For each case compute the time rate of change of the speed.

I really wasn't sure how to do the problem, so I sort of just guessed at a lot of things. So any help would be great. But this is what I did.

I assumed the only acceleration after the ball was thrown was due to gravity.
so I got
a_n=-32.2*cos(30) a_t=-32.2*sin(30)
then I used a_n=\frac{v^2}{\rho} to solve for the curvature \rho for when the ball was just released I got \rho = 358.6 ft . Then I used \beta'=\frac{100}{\rho} to solve for the angular rate.

I wasnt sure on how to solve (b) so I just assumed that the angle between \theta and the x- axis would be 0 degrees since the apex is the highest point and I guessed that would be when the ball would stop moving up and start going down.

So I got
a_n = -32,2 ft/s^2
then used
a_n=v*\beta' to solve for v at the apex using the the angular rate i got earlier for beta'.
I got v=115.4ft/s
then using the v and a_n I solved for \rho using the equation a_n=\frac{v^2}{\rho}
I got \rho= 413 ft

I'm definitely not confident with this answer, it's the first time I've done a problem like this, and I'm not sure if everything I did and assumed was legal.
Thanks in advance for any help.

 

[wais]

Hi there! Thank you for reaching out for help with this problem. Let's go through it step by step to make sure we understand the concepts and calculations involved.

Firstly, it is correct to assume that the only acceleration acting on the ball after it is released is due to gravity, as there are no other external forces acting on it.

Next, let's define our coordinate system. We can use a Cartesian coordinate system, with the x-axis being the horizontal direction and the y-axis being the vertical direction. The initial velocity, v_0, can be broken down into its x and y components as v_{0x}=v_0cos(\theta) and v_{0y}=v_0sin(\theta).

(a) To determine the radius of curvature just after release, we can use the equation a_n=\frac{v^2}{\rho}, where a_n is the normal acceleration and \rho is the radius of curvature. We can substitute in the values we know: a_n=-g=-32.2 ft/s^2 and v=v_0=100 ft/s. Solving for \rho, we get \rho=\frac{v^2}{a_n}=\frac{100^2}{32.2}=310.56 ft.

To find the time rate of change of speed, we can use the equation a_t=\frac{dv}{dt}, where a_t is the tangential acceleration. Since there is no tangential acceleration in this case, the time rate of change of speed is 0 ft/s^2.

(b) To determine the radius of curvature at the apex, we can use the same equation as before, but with the values at the apex instead. At the apex, the velocity will be purely horizontal, so v=v_{apex}=v_{0x}=v_0cos(\theta)=86.6 ft/s. Using the same equation, we get a_n=\frac{v^2}{\rho}=-g=-32.2 ft/s^2. Solving for \rho, we get \rho=\frac{v^2}{a_n}=\frac{86.6^2}{32.2}=232.48 ft.

To find the time rate of change of speed at the apex, we can use the same equation as before, but with the tangential acceleration being due to the change in vertical velocity. As the ball reaches the apex, its vertical velocity