Projectile Motion from Ramp
Date: Fall 2012
A 2-kg solid ball rolls down an incline in which the vertical drop is 30 m. At the bottom of the incline there is a horizontal length of track the ball travels before it reaches the end and falls to the ground 10 m below. How for horizontally does the ball travel after it leaves the track until it hits the ground? Ignore friction. According to the key the answer is 29.3 m.
I need help with the velocity of the ball as it leaves the ramp. I incorrectly assumed the velocity would be the same as if the ball were dropped and used v=the square root of (2ad). Which gives me 24.26m/s. If I use v=the square root of (sin 45 (2ad)) then I get the correct answer. Is this because some of the potential energy was used for horizontal motion? and Why 45 degrees?
I get 34.6m, from a fall time of sqrt(20/a). I also assumed the velocity of the ball was sqrt(2ad) as it left the track. It may be that we were supposed to figure out how much of the velocity at the end of the incline was vertical and how much was horizontal, but we do not have enough information (incline angle) to do that. Also, if we were to decompose the velocity into components, it would have been something like sin(angle)*sqrt(2ad), not sqrt(sin(angle)*2ad).
It is reasonable to assume that all of the potential energy is accounted for in the horizontal velocity of the ball. If the transition from inclined to horizontal were smooth, with a radius of curvature larger than the radius of the ball, then the force redirecting the ball would be everywhere normal to the ball's velocity, so it could not slow the ball down. That is not the picture we are given, but without the incline angle, it is the only assumption that yields a solvable problem.
The explanation for the 10/7 is rotational kinetic energy. When the
question says neglect friction, it apparently *doesn't* mean to
neglect the friction that causes the ball to roll. (when I first did
the problem, I noted that the ball would slide, not roll, if you
neglect friction...but anyway...)
K.E. due to linear motion + K.E. due to rotation = potential energy
1/2 mv^2 + 1/2 I * O^2 =mgh
with O=omega=v/r; I=2/5 m r^2 (moment of inertia for a solid sphere)
1/2 mv^2 +1/2(v^2/r^2*2/5*m*r^2)
(1/2 + 1/2*2/5) m v^2 = mgh
So if you "sorta" neglect friction you get that answer... if you
entirely neglect friction (e.g. the ball slides instead of rolls),
then you get a different answer.
(Credit to my physicist friend who pointed this out to me...)
Hope this helps,
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Update: November 2011