What is the period of this pendulum if the elevator accelerates upward with an acceleration A?

PHY132         Exam 1: Ch 13 – 15

First derive the expression to calculate the work required to move a satellite from Earth to orbit.  Then specifically derive an expression for the work required to move an Earth satellite of mass m from a circular orbit of radius 3RE to one of radius 4RE.

A simple pendulum is 4.00 m long. (a) What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 6.00 m/s2? (b) What is its period if the elevator is accelerating downward at 6.00 m/s2? (c) What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 6.00 m/s2?

            a. (a) 6.45 s; (b) 3.16 s; (c) 3.71 s                           b. (a) 3.16 s; (b) 6.45 s; (c) 4.01 s

            c. (a) 3.16 s; (b) 6.45 s; (c) 3.71 s                            d. (a) 6.45 s; (b) 3.16 s; (c) 4.01 s

T = 2π/ω = 2π/√(g/L)
(a) T = 2π/√(g+a /4) T = 2π/√(10+6 / 4) T = 3.14 sec	(b) T = 2π/√(g-a /4) T = 2π/√(10-6 / 4) T = 6.28 sec	(c) T = 2π/√((g2+a2)1/2 /4) T = 2π/√( 102+62)1/2 / 4) T = 3.68 sec

A frog in a hemispherical bowl of 5 cm radius and negligible mass finds that he floats without sinking into a sea of blue-green ooze having a density of 1.50 g/cm3.  Then a steel radio-transmitter (mtransmitter = 30 g) is attached to the frog.  Mercury is added to the blue-green ooze to increase the density.  What is the new density of the Mercury laden blue-green ooze to support the new mass?
mg = FB à weight of displace fluid mg = mFluidg   ρ = m/V mg = ρgV m =        ρ        V m = 1.5 g/cc * ½ (4π/3 * 53) m = 1.5 g/cc * 262 cc m = 393 g	mnew = 393 + 30g ρ =    m  /   V ρ = 423 / 262 ρ = 1.61 g/cc

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Imagine that you are in a stationary elevator and you drop an object.
Using a distance measuring scale attached to the elevator and a timing device you measure the acceleration of free fall relative to the elevator as $g$.

Now repeat the experiment in an elevator which is moving up or down at constant velocity relative to the Earth.
Again using the scale attached to the elevator and a timing device you measure the acceleration of free fall relative to the elevator as $g$.

Next repeat the experiment in an elevator accelerating upwards at an acceleration $a$ relative to the Earth and you find that the acceleration of free fall measured relative to the elevator is $g+a$. To show that this is reasonable repeat the experiment with the elevator in free fall relative to the Earth ie accelerating downwards with an acceleration of $g$ relative to the Earth.

If you "drop" an object in such a situation and you are in the elevator you will not see the object moving relative to you or the elevator as you, the elevator and the object are all accelerating downwards at $g$ relative to the earth.

What does all this tell you? It tells you that as far as an observer in an accelerating, relative to the Earth, elevator is concerned the acceleration of free fall is $g + (\pm a)$ where $\pm a$ is the acceleration of the elevator relative to the Earth with the positive sign for an upward acceleration relative to the Earth and the negative sign for a downward acceleration relative to the Earth.

Note that if the downward acceleration of the elevator relative to the Earth is greater than $g$ you have a situation where a "dropped" object actually accelerates upwards relative to the elevator.

Now consider your question as to whether one should add or subtract $a$ from $g$ in a situation where the elevator is in free fall relative to the Earth. You, standing in the elevator, pull the pendulum bob from its equilibrium position and release it. The pendulum bob does not move relative to you or the elevator. The period of the bob is "infinite" because the tension in the string which is connecting the bob to the elevator is zero.

Accelerate the elevator at $2g$ downwards relative to the Earth and you will measure the same period for the pendulum as in a stationary elevator with the pendulum oscillating "up side down" ie the point of suspension will be below the pendulum bob.

In your equation for the period of the pendulum the $g$ is the value of the acceleration of free fall of a body relative to the elevator. and when the elevator is accelerating upwards with an acceleration of $a$ relative to the Earth the acceleration of free fall relative to the elevator is $g \; { \Large +} \;a$.

You might find it interesting to read about the Equivalence Principle?