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Sergio039 [100]

3 years ago

7

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

he acceleration of gravity is 9.8 m/s 2 . What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on

1 answer:

tekilochka [14]3 years ago

7 0

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

<u>ω = 0.05 rad/s</u>

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An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

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Suppose after t sec of throw camera reach balloon then,

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$s=3\times t$

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$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

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$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

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