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3 years ago

6

An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch p

ad takes 1.25 s for the same cycle of swing during liftoff.
What is the acceleration (m/s²) of the rocket?(Hint: Inside the rocket, it appears that g has increased.)

1 answer:

klio [65]3 years ago

5 0

Answer:

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π. $\sqrt{\frac{L}{a} }$

where

L = length
a = acceleration

therefore the ration of the periods while on ground and at take off will be

$\frac{T1}{T2}$ =(2π $\sqrt{\frac{L}{a1} }$ ) / (2π $\sqrt{\frac{L}{a2} }$ )

where

a1 = acceleration on ground while waiting
a2 = acceleration during liftoff

$\frac{T1}{T2}$ = $\frac{\sqrt{\frac{L}{a1} }}{\sqrt{\frac{L}{a2} }}$

squaring both sides we have

$(\frac{T1}{T2})^{2}$ = $\frac{\frac{L}{a1} }{\frac{L}{a2} }$

$(\frac{T1}{T2})^{2}$ = $\frac{a2}{a1}$

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

$(\frac{T1}{T2})^{2}$ = $\frac{a2}{9.8}$

a2 = 9.8 x $(\frac{T1}{T2})^{2}$

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x $(\frac{2.45}{1.25})^{2}$

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.

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