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3 years ago

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Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and

diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of a bucket of water is 20 kg. Someone cranks the bucket up and then lets go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?

1 answer:

sleet_krkn [62]3 years ago

4 0

Answer:

$\alpha=78.4\ rad.s^{-2}$

Explanation:

Given:

mass of solid cylinder, $m=50\ kg$

diameter of cylinder, $d=0.25\ m$

mass of bucket of water, $m_w=20\ kg$

<em>When the bucket is released to fall in the well, it fall under the acceleration due to gravity.</em>

We have formula for angular acceleration as:

$\alpha=\frac{g}{r}$

where:

g = acceleration due to gravity

r = radius of the cylinder

$\aplha=\frac{9.8}{0.125}$

$\alpha=78.4\ rad.s^{-2}$

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Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

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here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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P=IV
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Answer:

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Answer:

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Answer:

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