I wish the Earth wasn't rotating

If a crate of mass 12 k g 12kg is placed on a weighing machine at a place on earth then calculate the measured weight in SI unit if the rotation of the earth has to be taken into account.

Details and assumption

  • The acceleration due to gravity ( a g ) = 9.806 (a_g)=9.806 S . I . u n i t S.I. unit .
  • The angular velocity of the earth ( ω ) = 7.292 × 1 0 5 (\omega)=7.292\times10^{-5} S . I . u n i t S.I. unit .
  • The radius of earth ( r ) = 6.371 × 1 0 6 (r)=6.371\times10^6 S . I . u n i t S.I. unit .

If your answer is x x , then enter your answer as x × 1 0 3 x\times10^3 and round it off to the nearest integer


The answer is 117265.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

In this solution, I have considered the direction towards the outward from the centre of the earth as positive.

  • The normal force ( F n ) (F_n) on the crate from the scale is directed outward, in the positive direction of r-axis.

  • The gravitational force, represented with its equivalent m a g ma_g , is directed inward(negative of r-axis). Because it travels in a circle about the center of the earth, as the earth turns the crate has a centripetal acceleration a r = ω 2 r a_r=\omega^2r directed toward the earth's centre(negative of r-axis)

Thus F n e t , r = m a r F_{net,r}=ma_r

F n m a g = m ( ω 2 r ) \Rightarrow F_n-ma_g=m(-\omega^2r)

m g = m ( a g ω 2 r ) \Rightarrow mg=m(a_g-\omega^2r)

Putting the values given in question, m g = 117.265 N mg=117.265N

So the answer is 117265 \boxed{117265}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...