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3 years ago

7

A peg is located a distance h directly below the point of attachment of the cord. If h = 0.760 L, what will be the speed of the

ball when it reaches the top of its circular path about the peg?

1 answer:

Rufina [12.5K]3 years ago

6 0

Answer:

Explanation:

Given

Pivot is at h=0.76 L

Where L is the length of String

Conserving Energy at A and B

$mgL=\frac{1}{2}mu^2$

where u=velocity at bottom

$u=\sqrt{2gL}$

After coming at bottom the ball completes the circle with radius r=L-0.76 L

Suppose v is the velocity at the top

Conserving Energy at B and C

$\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2$

Eliminating m

$u^2=4r+v^2$

$v^2=u^2-4\cdot gr$

$v^2=2gL-4g(L-0.76L)$

$v^2=1.04gL$

$v=\sqrt{1.04gL}$

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Answer:

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When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

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Answer:

The correct answer is

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The transformation equation fom potential to kinetic energy is =

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