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4 years ago

9

A 500g ball moves in a vertical circle on a 102cm long string. If the speed at the top is 4.0m/s, then the speed at the bottom w

ill be 7.5m/s.
A) What is the ball's weight?

B) What is the tension in the string when the ball is at the top?

C) What is the tension in the string when the ball is at the bottom?

1 answer:

Len [333]4 years ago

3 0

Answer:

Explanation:

ball's weight = mg = .5 x 9.8 = 4.9 N.

If T₁ be tension at the top , centripetal force provided at the top

= mg + T₁ = 4.9 + T₁

4.9 + T₁ = m v² / r , v = velocity at the top , r = radius of circular path = 1.02 m

4.9 + T₁ = .5 x 4² / 1.02

4.9 + T₁ = .5 x 4² / 1.02

= T₁ = 7.843 - 4.9 = 2.943 N

If T₂ be tension at the bottom , centripetal force provided at the bottom

= T₂ - mg = T₂ - 4.9

T₂ - 4.9 = m v² / r , v = velocity at the bottom , r = radius of circular path = 1.02 m

T₂ - 4.9= .5 x 7.5² / 1.02

T₂ - 4.9 = .5 x 7.5² / 1.02

T₂ = 32.47 N

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Explanation:

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We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

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B)

The kinetic energy of an object is given by

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In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

D)

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

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F)

The angular momentum of an object is defined as

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First, we rewrite the list of expressions for the different quantities that we found:

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