- #1
mohabitar
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We're supposed to analyze this video:
https://www.youtube.com/watch?v=MMMkENFOysQ
Givens
3) For clarity, I shall name the idiot that stayed in the merry-go-round, Chuck, and the idiot that lost his grip, Norris.
4) Norris impacted the Earth approximately 3.5 m away from the point he let go and “g” is -9.81m/s2 as always.
5) Only Norris’s uniform circular motion contributed to his velocity prior to letting go.
6) The diameter of the merry go round is 2.75m and its height is 1.15m.
7) Norris let's go at Time t=0s.
7) Do not worry about the fact that the actual circular motion in the video is not uniform. We will ignore the fact that the rotation of merry-go-round is accelerating.
Also,
V is a) [parallel] to the ground and b)[tangent] to the merry-go-round.
_______________________
Questions:
2) At t=0, what is the magnitude of Norris’s velocity vector with respect to the ground?
______
So here's the answer to that question:
Treating Norris as a projectile, whose Dx is 3.5m and Dy is 1.15m, it can be determined from the givens and the answer to 1 that his initial velocity is [7.23 m/s**2]
So I've been trying to replicate this answer but with no success. I'm not really sure what formula to use. So I've tried different cases with different assumptions.
-So his final velocity is 0, since he lands. I've tried using v^2=u^2+2as to figure out u(initial velocity). For a, I used gravity, but it could be the centripetal acceleration, which we would need velocity to find anyway, so that couldn't be it. For s, which is the displacement, I tried using 3.5 as the horizontal displacement, I tried adding 1.15 to that, I tried using pythagorean to find hypotenuse-but none of those gave me that answer. I really have no idea what to do here.
https://www.youtube.com/watch?v=MMMkENFOysQ
Givens
3) For clarity, I shall name the idiot that stayed in the merry-go-round, Chuck, and the idiot that lost his grip, Norris.
4) Norris impacted the Earth approximately 3.5 m away from the point he let go and “g” is -9.81m/s2 as always.
5) Only Norris’s uniform circular motion contributed to his velocity prior to letting go.
6) The diameter of the merry go round is 2.75m and its height is 1.15m.
7) Norris let's go at Time t=0s.
7) Do not worry about the fact that the actual circular motion in the video is not uniform. We will ignore the fact that the rotation of merry-go-round is accelerating.
Also,
V is a) [parallel] to the ground and b)[tangent] to the merry-go-round.
_______________________
Questions:
2) At t=0, what is the magnitude of Norris’s velocity vector with respect to the ground?
______
So here's the answer to that question:
Treating Norris as a projectile, whose Dx is 3.5m and Dy is 1.15m, it can be determined from the givens and the answer to 1 that his initial velocity is [7.23 m/s**2]
So I've been trying to replicate this answer but with no success. I'm not really sure what formula to use. So I've tried different cases with different assumptions.
-So his final velocity is 0, since he lands. I've tried using v^2=u^2+2as to figure out u(initial velocity). For a, I used gravity, but it could be the centripetal acceleration, which we would need velocity to find anyway, so that couldn't be it. For s, which is the displacement, I tried using 3.5 as the horizontal displacement, I tried adding 1.15 to that, I tried using pythagorean to find hypotenuse-but none of those gave me that answer. I really have no idea what to do here.
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