# Submarine Pendulum

A scientist is making a precise measurement of g at a certain point in the Indian Ocean (on the equator) by timing the swing of a pendulum of accurately known construction. To provide a stable base, the measurements are conducted in a submerged submarine. It is observed that a slightly different value for g is obtained when the submarine is in motion. What is the change in g, (g'-g) in mm/s2, if the submarine is moving east at 11.3 kph? (Include the sign of the change in your answer.) Use as the equatorial radius of the Earth 6378.2 km

i have no clue how to do this any help would be appreciated

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Doc Al
Mentor
Hint: What kind of motion does the submarine (and pendulum) undergo?

it undergoes centripcal acceleration but the equation for that is v^2/r

or do you mean constant motion im so lost for this question

it's centripedal ACCELERATION. Constant velocity will have no effect on the pendulum nor on g.

Doc Al
Mentor
it undergoes centripcal acceleration but the equation for that is v^2/r
Yes, the submarine is centripetally accelerated. Now use that fact to figure out the change in the measured value of g.

so take the speed we right now are going because of the earth's rotation and divide it by the radius then add the speed of the submarine to the speed were going right now and find the difference between them

no that wont work, i still dont know how to use the fact that it is centripedal acceleration to my advantage (a=v^2/r) using that doesn't give me a value close to g
or anything close to 9.81 to compare the subs accerations to

Doc Al
Mentor
Note that the problem says: "It is observed that a slightly different value for g is obtained when the submarine is in motion."

How does the centripetal acceleration change when the submarine moves?

it changes slightly but how do you calculate that

Doc Al
Mentor
Hint: When the submarine is "at rest" in the water it is moving at the same rate as the rotating earth.

thanks for the help i got the answer last night, thank-you