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3 years ago

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all bearings are made by lebng spherical drops of molten metal fall inside a tall tower – called a shot tower – and solidify as

they fall a height h. The bearings are released from rest. a)Draw the Problem. b)If a bearing needs 4.0 s to solidify enough for impact, how high must the tower be? (LC) c)What is the bearing’s impact velocity? (LC)

1 answer:

Wewaii [24]3 years ago

7 0

Answer:

Part b)

h = 78.5 m

Part c)

v = 39.24 m/s

Explanation:

Part b)

If ball need t = 0 to t = 4 s then height of the tower is the total displacement of the ball in t = 4 s interval

here if ball start from rest

then its displacement is given as

$\Delta y = \frac{1}{2}gt^2$

$\Delta y = \frac{1}{2}(9.81)(4^2)$

$\Delta y = 78.5 m$

Part c)

Speed of the bearing at the end of the motion of the ball

$v_f = v_i + at$

$v_f = 0 + (9.81)(4)$

$v_f = 39.24 m/s$

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The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

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Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

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