Viscous forces under freefall


by Binayak95
Tags: forces, freefall, viscous
Binayak95
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#1
May3-12, 08:56 AM
P: 5
A sphere inside an elevator is falling at terminal velocity. Say the elevator cables break and it starts to freefall, what happens to the sphere's velocity????

I have developed some rudimentary solution but would like to see a confirmation.
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Khashishi
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#2
May3-12, 10:30 AM
P: 833
Terminal velocity is due to air resistance equal to the gravitational force. When the elevator starts moving, the air in the elevator will be jostled along with the elevator car, and the sphere will be jostled along with the air. The exact motion will be messy and far from an ideal situation that you usually see in gedanken experiments.
boneh3ad
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#3
May3-12, 12:58 PM
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The terminal velocity with respect to an inertial frame would increase in the freefalling elevator compared to a stationary one or one that is moving upward slowly.

It will remain the same with respect to the air in the elevator.

Rear Naked
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#4
May3-12, 03:22 PM
P: 48

Viscous forces under freefall


Interesting problem. I'm too dumb to answer.
Binayak95
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#5
May4-12, 01:55 AM
P: 5
I thought that since g would drop to zero bouyant force would also reduce to zero. Since Viscous force would still exist, the sphere's velocity would reduce and eventually the sphere would be pushed against the elevator's roof.
boneh3ad
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#6
May4-12, 08:19 AM
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The buoyant force in air is essentially negligible except for very large, light objects like zeppelins. Compared to other forces at work, it is almost always very small.

Now, since your elevator is moving down, the air inside is moving down with the same speed with regard to an inertial frame (say, the elevator shaft). Viscous forces depend on the speed of the object with respect to the fluid it is traveling through, so if you had one ball falling inside the elevator and one ball falling outside of it through the shaft with the same initial speed (with respect to the shaft), the ball outside the elevator would be moving with a greater speed with respect to the fluid it is moving through than the one inside the elevator, so viscous forces would be greater on the ball in the shaft rather than the ball in the elevator.
Khashishi
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#7
May4-12, 09:17 AM
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Quote Quote by Binayak95 View Post
I thought that since g would drop to zero bouyant force would also reduce to zero. Since Viscous force would still exist, the sphere's velocity would reduce and eventually the sphere would be pushed against the elevator's roof.
No, that's wrong. After some time (long elevator shaft), the sphere's velocity will go to zero relative to the elevator, but it won't be pushed against the root.
boneh3ad
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#8
May4-12, 10:30 AM
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Quote Quote by Khashishi View Post
No, that's wrong. After some time (long elevator shaft), the sphere's velocity will go to zero relative to the elevator, but it won't be pushed against the root.
That is also not true unless the ball was neutrally buoyant in the elevator or the elevator is also in free fall. If the elevator was also in free fall and the ball was released at the same time, the ball would remain at the same velocity as the elevator if it was a vacuum, but because of air, the ball will still fall faster since it will have less drag on it than the elevator and so will accelerate more.


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