Question

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?

Answer 1

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

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

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

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

$v^2=1.04gL$

$v=\sqrt{1.04gL}$