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cupoosta [38]
3 years ago
13

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?
Physics
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:

t_1 = \frac{v_i}{a_i}

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As t_1 = t_2 we can simplify s(t) = v_it_1

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3 years ago
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Answer:

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Answer:

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