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fenix001 [56]
3 years ago
8

A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio

n in the cord connected to the ball be greatest?
a. A little after the bottom of the circle when the ball is climbing.
b. A little before the bottom of the circle when the ball is descending quickly.
c. At the bottom of the circle.
d. Nowhere; the cord is stretched the same amount at all points.
e. At the top of the circle.

Physics
1 answer:
BARSIC [14]3 years ago
8 0

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T =  m ω² r  + m g

<h3>At the top condition</h3>

T + m g = m a

T=  m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

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D₂= 167,21 cm : Magnitude  of the second displacement

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We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

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R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

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D₂y=142.83-80.65=62.18 cm

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D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

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β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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