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NISA [10]
3 years ago
15

It takes the Earth 24 hours to make a complete rotation around its axis. 33% Part (a) What is the period of rotation of the Eart

h in seconds? Grade Summary Deductions0 Potential 100 sin cotan) coso asin) atano acotan) sinh0 tan) acoso Attempts remaining per antemp4) detailed vie cosh0 tanhO cotanhO ODegrees O Radians Hint I give up Hints:-% deduction per hint. Hints remaining:- Feedback: dohction per fcedback ▲ 33% Part (b) What is the angular velocity of the Earth in rad/s? 33% Part (c) Given that Earth has a radius of 6.4 x 106 m at its equator, what is the linear velocity at Earth's surface?
Physics
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

Explanation:

a )

one rotation per 24 hours

Time period of rotation of the earth T =  time / no of rotation

T = 24 x 60 x 60 s / 1

= 86400 s .

b )

angular velocity = angle of rotation in radian / time

in one rotation , angle made at the centre = 2π radian

= 2 x 3.14 radian

angular velocity ω = 2π / T

= 2 x 3.14 / 86400  radian / s

= 72.68 x 10⁻⁶ radian / s

c )

Relation between linear and angular velocity is as follows

v = ω  x R where R is radius of the earth and v is linear velocity .

linear velocity = 72.68 x 10⁻⁶ x 6.4 x 10⁶ m /s

= 465.152 m /s

=

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Answer:

μsmín = 0.1

Explanation:

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       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

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  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

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      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

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  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

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