A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just before it hits the ground? (Assume the lower end of the pole does not slip.)

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aristocles aristocles · Answer 1 · 2019-08-01T02:40:16+02:00

Answer:

v = 17.66 m/s

Explanation:

As we know that the lower end of the pole is fixed in the ground and it start rotating about that end

so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end

so we have

[tex]mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2[/tex]

so we have

[tex]\omega = \sqrt{\frac{6g}{L}}[/tex]

now we have

[tex]\omega = \sqrt{\frac{6(9.81)}{5.30}}[/tex]

[tex]\omega = 3.33 rad/s[/tex]

now the speed of the other tip of the pole is given as

[tex]v = \omega L[/tex]

[tex]v = (3.33)(5.30) = 17.66 m/s[/tex]