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3 years ago

What is the angular velocity of a 6–foot pendulum that takes 3 seconds to complete an arc of 14.13 feet? Use 3.14 for π

1 answer:

3 0

1) Calculate the whole arc (complete circumference):

Circumference = 2π*radius

radius = 6 foot

=> Circumference = 2 * 3.14 * 6 foot = 37.68

2) Calculate the time to make a whole round, by making a proportion with the two ratios:

ratio 1 = x / 37.68 =

ratio 2 = 3 / 14.13

ratio 1 = ratio 2 =>

   x 3s
---------- =   -----------
37.68 ft    14.13 ft

=> x = 3 * 37.68 / 14.13 = 8 s

3) Calcuate the angular velocity as the 2π rad (which is the total angle of a circle) divided by the time.

So the angular velocity is 2π rad / 8 s = 2*3.14 foot / 8 s = 0.785 rad / s

Answer: 0.785 rad / s

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) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

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$x=0,y=0\implies z=2\implies(0,0,2)$

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$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

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3 years ago