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katrin2010 [14]

2 years ago

6

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

h a period of 24 hours. Consider only the effect of this rotation on the person on the surface. (Ignore Earth's orbital motion about the Sun.) (a) What is the speed and what is the magnitude of the acceleration of a person standing on the equator?

1 answer:

cestrela7 [59]2 years ago

6 0

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

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The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

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From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

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$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

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$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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