# Circular motion

## Homework Statement

A plane is travelling at 200 m/s following the arc of a vertical circle of radius R. At the top of its path, the passengers experience "weightlessness" .To one significant, what is the value of R ?

## Homework Equations

Mv^2/R=Mat , N=mg(1+at/g)
Velocity orbit = squareroot(RG)

## The Attempt at a Solution

I try to use the velocity or orbit to this question until I saw the word " weightlessness" . That means there is apparent weight is zero but how could that apply to this case ? Does that mean N , apparent weight - mg=Mat , at is tangential acceleration ? and I consider N = 0 in this case. Does that make sense ? so F net = N-Mg=Mat and N=Mg(1+at/g) but N=0 , so at = -g and plug in to mv^/r =mat ?

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HallsofIvy
Homework Helper
Without force, of course, an object would move in a straight line. In order to move in a circle at constant speed, there must be a "centripetal force" directed toward the center of the circle. What that force is depends upon both the speed and the radius of the circle. Of course, here there will also be gravitational force, mg, directed downward. The "apparent weight" is 0 when those two forces are equal and opposite.

CWatters
Homework Helper
Gold Member
Without force, of course, an object would move in a straight line. In order to move in a circle at constant speed, there must be a "centripetal force" directed toward the center of the circle. What that force is depends upon both the speed and the radius of the circle. Of course, here there will also be gravitational force, mg, directed downward. The "apparent weight" is 0 when those two forces are equal and opposite.
That's slightly confusing because both centripetal force and gravity act downwards :-)

CWatters
Homework Helper
Gold Member
To feel weightless you must be in free fall (eg accelerating downwards at a rate equivalent to g). That acceleration is provided by centripetal acceleration.

haruspex
Homework Helper
Gold Member
The "apparent weight" is 0 when those two forces are equal and opposite.
As CWatters says, that's confusing (well, wrong really).
Gravity is one of the actual forces applied, while centripetal force is a required resultant of the actual forces to achieve a known acceleration orthogonal to the velocity. It's cause versus affect. Apparent weight is zero when there are no other forces acting in the direction of gravity, which means that gravity and the vertical component of centripetal force are equal and in the same direction.

PeterO
Homework Helper

## Homework Statement

A plane is travelling at 200 m/s following the arc of a vertical circle of radius R. At the top of its path, the passengers experience "weightlessness" .To one significant, what is the value of R ?

## Homework Equations

Mv^2/R=Mat , N=mg(1+at/g)
Velocity orbit = squareroot(RG)

## The Attempt at a Solution

I try to use the velocity or orbit to this question until I saw the word " weightlessness" . That means there is apparent weight is zero but how could that apply to this case ? Does that mean N , apparent weight - mg=Mat , at is tangential acceleration ? and I consider N = 0 in this case. Does that make sense ? so F net = N-Mg=Mat and N=Mg(1+at/g) but N=0 , so at = -g and plug in to mv^/r =mat ?
This question could have been set as a standard Ferris Wheel / amusement park ride - except that you would probably object to a Ferris wheel rotating at 200 m/s (720 kph), so you were told it was an aeroplane.
It still amounts to the same method of analysis.