Circular Motion Homework: Plane at 200m/s, R Radius to Zero Weight

AI Thread Summary
A plane traveling at 200 m/s in a vertical circular path creates a scenario where passengers experience weightlessness at the top due to the balance of gravitational and centripetal forces. The apparent weight is zero when the centripetal force equals the gravitational force, leading to the equation N - mg = Mat, where N is the normal force. The discussion highlights confusion about the roles of centripetal force and gravity, clarifying that apparent weightlessness occurs when these forces are equal and opposite. The problem can be likened to a Ferris wheel scenario, emphasizing the need for proper analysis of forces involved. Understanding these dynamics is crucial for solving for the radius R in this context.
Anthonyphy2013
Messages
30
Reaction score
0

Homework Statement



A plane is traveling 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 into mv^/r =mat ?
 
Physics news on Phys.org
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.
 
HallsofIvy said:
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 :-)
 
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.
 
HallsofIvy said:
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.
 
Anthonyphy2013 said:

Homework Statement



A plane is traveling 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 into 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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top