If you lose your grip on a rapidly spinning merry-go-round and fall off, in which direction would you fly? Explain.
2007-11-03 09:56:22 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
On a rapidly spinning merry-go-round, your velocity is a constantly changing vector that is tangent to the circle you are moving around. This changing velocity is caused by the centripetal force exerted by the merry-go-round on your body.
If you are hanging on for dear life, your legs and feet pointing outward away from the axis of rotation, after you let go there are no forces acting on your body. Your velocity vector is unchanged. You go flying off in a direction that is tangent to the rotation circle of the merry-go-round, in the same direction as the rotation. In other words, you go flying off in the last direction your velocity vector was pointing when you let go.
2007-11-03 10:21:13 · answer #1 · answered by hevans1944 5 · 0⤊ 0⤋
you would continue on in a straight line from the point that you lost your grip. The centripetal force is that force, that pull between you and the center of the merry-go-round, so one that force is broken by you losing your grip, your velocity will carry you in a straight line with 3 points. Point a would be one end of the centripetal force, the center of the merry-go-round, point b would be where you lost your grip and point c would be where you land, all in a straight line but the line would change as you rotate around the center and the velocity at which you are traveling will bend the outer end of that line in the direction you were traveling because of your velocity.
the best that I can do explaining that but if you spin a bucket around you on a rope, the tensions on the rope would be the centripetal force between you and the bucket and if you then release the rope, the bucket will continue in the direction your hand was facing when the centripetal force was broken by you releasing the rope.
2007-11-03 10:30:38 · answer #2 · answered by Al B 7 · 0⤊ 0⤋
90 degree to the direction of circular motion of that instant. As bcos you are moving forward (a force is in that way) and a force is also towards the centre. When the grip is gone(!!) the forward force will act and u'll be thrown at right angle to the force that was previously towards the centre.
2007-11-03 10:04:44 · answer #3 · answered by Psyche 2 · 0⤊ 1⤋
the object while rotating around your hand is formind a circle, the moment you lose grip on the rotating object , it has no force on it ... so it will keep in the same velocity and the same direction
I mean it will continue to go with the tangent of the radious of the point in which the rotating object was while rotating.
those may help you understand :
http://en.wikipedia.org/wiki/Centripetal_force
http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html
2007-11-03 10:05:02 · answer #4 · answered by wise MONKEY 2 · 0⤊ 0⤋
To save on with a around course the acceleration in the direction of the middle of the music could desire to = v^2/r, so the rigidity could desire to = mv^2/eighty. the optimum frictional rigidity attainable = established rigidity circumstances coefficient of friction (by the way, coefficient of friction does not have a unit. do no longer write it with the N). the line isn't banked, so the conventional rigidity is equivalent in value to the gravitational rigidity. Frictional rigidity ? (0.30) m (9.8 m/s^2) to come across the optimum v, set the values equivalent mv^2/80m = (0.30) m (9.8 m/s^2) m cancels, and v = ?((0.30)(9.8)(eighty)) , slightly over 15m/s
2016-10-03 06:42:12 · answer #5 · answered by syverson 4 · 0⤊ 0⤋
Away from the centipede.
2007-11-03 10:00:06 · answer #6 · answered by calm down now 5 · 0⤊ 1⤋
clockwise
2007-11-03 09:58:14 · answer #7 · answered by Mr Answers 2 · 0⤊ 1⤋