Dynamics: Normal and Tangential projectile motion problem

[ajt49]

Homework Statement

A canon fires a projectile at a speed of 100 m/s at angle of 20 degrees. (a) what is the radius of curvature of the projectile at its highest altitude? (b) What is the radius of curvature of the path .5 seconds after firing?

Homework Equations

Velocity vector = s'u_r

Acceleration vector = v' =
a_tu_t + a_nu_n

where,

a_t= v'= dv/dt

&

a_n= [v^(2)]/p

The Attempt at a Solution

Wrote out the given which was,

Vo = 100 m/s at an angle 20 degrees above the horizontal x⁺ axis

Vo_x= Vo*cos(20)

Vo_y= Vo*sin(20)

a_x = 0 = a_t = 0

a_y= g or -9.81 acting in the negative j direction.

(a) a_n = -9.81 ==> -9.81 = [v^2]/p ==> p = [100^2]/9.81 = 1019.37 meters

 

[BruceW]

hey! welcome to physicsforums :)

Uh, your answer for part a is almost right. But v is not 100m/s. Because 100m/s is the initial velocity. So what velocity does the object have when at the highest altitude?

 

[ajt49]

Thanks for the welcome and help!

Well at the highest altitude I guess the y component of the velocity is zero, so I would only have to calculate the x component of velocity. Therefore, a_n= [v^2]/p is actually

-9.81=[(100m/s*cos 20)^2]/p .. solve for row and p= 900.12 meters? Would I use the equations (V_osin theta)t and (V_ocos theta)t at time .5 seconds? the are vectors so I could find my speed by Pythagorean theorem.

 

[BruceW]

yes, I agree with your answer of 900.12m for part a. And for part b, yes, that sounds like a good plan. find the vector velocity at that time, and then you can find both a_n and |v|. (remember that a_n is not the total acceleration, but the acceleration in the direction perpendicular to the velocity).

 

[HalfLostAndSearching]

(b) t = .5 seconds

Vx = Vox = 100*cos(20) = 94.28 m/s

Vy = Voy + at = 100*sin(20) - 9.81*.5 = -4.905 m/s

V= sqrt[94.28^2 + (-4.905)^2] = 94.3 m/s

an = [94.3^2]/p = -9.81 ==> p = [94.3^2]/(-9.81) = 899.6 meters

(a) The radius of curvature at the highest altitude would be equal to the radius of the circular path that the projectile is following at that point. This can be calculated using the formula for centripetal acceleration, which is given by an = v^2/r, where v is the velocity of the object and r is the radius of curvature. In this case, we can use the given value of an = -9.81 m/s^2 and the velocity of the projectile at its highest altitude, which is equal to the horizontal component of the initial velocity, Vx = 94.28 m/s. This gives us a radius of curvature of 954.2 meters.

(b) To calculate the radius of curvature at t = 0.5 seconds, we can use the same formula an = v^2/r, but this time we need to use the total velocity of the projectile, which is equal to 94.3 m/s. This gives us a radius of curvature of 899.6 meters. This is slightly smaller than the radius of curvature at the highest altitude, which makes sense since the projectile is accelerating downwards due to gravity.