Simple pendulum falling freely

[utkarshakash]

Homework Statement

The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will

a)continue its oscillation as before
b)stop
c)will go in a circular path
d)move on a straight line

Homework Equations

The Attempt at a Solution

When the box starts falling freely g effective is 0 as seen inside the box. That means the time period of oscillation tends to infinity. It means the pendulum must stop. But that's incorrect. The correct option is c).

 

[jedishrfu]

yes c seems right, at the bottom of the swing the velocity is tangential and the string is taut so with no gravity the string will restrain it forcing it to travel in a circular path much as you swinging a bucket around in a circle until it hits the ceiling.

 

[ehild]

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The Attempt at a Solution

When the box starts falling freely g effective is 0 as seen inside the box. That means the time period of oscillation tends to infinity. It means the pendulum must stop. But that's incorrect. The correct option is c).

If the box is in free fall you feel weightless in the box, as if you were in empty space. Is there any equilibrium point and restoring force for the bob? Can it be considered "pendulum"?
The bob of a pendulum moves forward and back along an arc of circle and it reaches its maximum displacement from the equilibrium point (and zero speed) during a quarter of the time period. Will the bob turn back when the box is in free fall? Will its initial speed decrease at all?

ehild

 

[lewando]

Agree with jedishfru. If the pendulum started free-falling when the bob was at its maximum height, momentarily motionless, it would remain that way as well, giving an infinite period. But at the lowest point, the bob has momentum.

 

[tomstringer]

Gravitational force ends but centripetal force, Fc = mv^2/L, remains where Fc is independent of g. Hence, the bob continues, in the absence of friction from the pivot, with constant velocity v.

Fnet = T + mgsin(Theta) = ma = mv^2/L = Fc where L is length of pendulum.
When g = 0, Fnet = Fc

 

[haruspex]

the bob continues, in the absence of friction from the pivot, with constant velocity v.

The question asks for its motion relative to the box. Since it is at a fixed distance from the pivot, that cannot be a straight line, so not constant velocity. If you meant constant velocity wrt an inertial frame, that would only be true if the box etc. is massless.

 

[tomstringer]

Isn't it true that the bob's velocity tangent to the circular path will become constant at the moment the box starts its freefall? I'm taking "relative to the box" to mean the same as "relative to the ceiling of the box". I see the bob continuing in its circular path at a constant speed, after the box's release, before hiting the ceiling.

 

[haruspex]

Isn't it true that the bob's velocity tangent to the circular path will become constant at the moment the box starts its freefall? I'm taking "relative to the box" to mean the same as "relative to the ceiling of the box". I see the bob continuing in its circular path at a constant speed, after the box's release, before hiting the ceiling.

You wrote 'constant velocity', which implies a straight line. Whether its tangential speed would be constant I'm not sure. Gets complicated when you consider the interaction with the mass of the box.

 

[gneill]

Since the problem doesn't state anything about the mass of the box (especially as compared to the mass of the pendulum), it's going to be hard to say anything precise about the motion of the pendulum with respect to the box, except that the two together will continue to follow the trajectory of the center of mass of the system at the instant of release. Oh, and angular momentum will be conserved...