A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.4 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 13.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?
Answer in m/s.....i have no clue how to solve this problem...i be greatly appriciate if someone can do it in step so i know how or where did that answer come from...thanks...
2007-10-15 05:11:53 · 3 answers · asked by pinkepromise 2 in Science & Mathematics ➔ Physics
The gain in kinetic energy of the ball is because of two factors:-
(1) As the ball comes down,its potential energy changes to kinetic energy
(2) Increase in speed due to force exerted by pitcher increases the kinetic energy
Total gain in KE=PE + F.s
Total gain in KE = mgh +F.( pi r)
Total gain in KE=0.250*g*2r + 33.0*pi*r
Total gain in KE=1.96 + 41.46 =43.42 J
Initial KE=(1/2)mu^2\
Initial KE=(1/2)0.250*13*13
Initial KE=21.125 J
Final KE=Initial KE + gain in KE
Final KE=21.125 + 43.42 =64.545 J
speed =sq rt [2* final KE /mass]
speed = sq rt [ 2* 64.545 / 0/250 ]
speed = sq rt 129.09/0.250
speed = sq rt 516.36 m/s
speed =22.72 m/s
The speed of the ball when released at the bottom is 22.72m/s
2007-10-15 07:31:50 · answer #1 · answered by ukmudgal 6 · 0⤊ 0⤋
Tough question.
You know the velocity at the top of the circle (13 m/sec) so what you need to know is how much the velocity has change by the time the ball is released at the bottom of the circle.
Two ways to solve this. If you have learned about Energy you will know that the change in energy is the Force * distance. The distance is the arc of the circle from top to the bottom (pi*r). You then solve the KE at the top, and the Energy and get the new KE and compute the new velocity from that.
Otherwise, you use your old standby F=ma to determine the aceleration and you use your equations delta (v)=a*t and d=1/2 at^2, remembering that d is pi*r and that you have an initial velocity. Much trickier to solve.
Good luck
2007-10-15 13:18:31 · answer #2 · answered by Frst Grade Rocks! Ω 7 · 0⤊ 0⤋
i'll show the method, you can plug in the numbers
lets assume gravity doesn't play a role in this problem.
1st step, figure out how much distance the ball travels
the length of an arc is given by L = r*theta.
since he's moving in a semi circle, theta = 180 degrees or pi radians. therefore,
L = .4*pi m
let's figure out the acceleration of the ball
from F = ma, a = F/m = 33/.250 m/s^2
from kinematics, vf^2 = vi^2 + 2*a*delta(s)
where delta(s) is the displacement which is L, so
vf^2 = vi^2 + 2*a*L
just plug in your numbers and presto!
2007-10-15 13:11:57 · answer #3 · answered by civil_av8r 7 · 0⤊ 0⤋