Question

A bucket containing water is tied to one end of a rope 5 m long and it is rotated in a
vertical circle about the other end. Find the number of rotations per minute in order

that the water in the bucket may not spill.​

Answer 1

Answer:

☆Concept:

▪︎object doing vertical circular motion will be under two existing forces:

1. gravity(mg) provided by the earth.
2. centripetal force ($\ m\omega^{2}r$) which is provided the tension of rope.

▪︎water in the bucket will press the surface, Normal reaction will be present for balancing.

▪︎due to centripetal force water will experience the outward centrifugal force.(with respect to moving object frame)

▪︎at the lowermost point water will experience maximum Normal reaction.

N = mg + $\bf{ m\omega^{2}r}$

▪︎at topmost point water will experience minimum to zero Normal,as the mg is almost balanced by the centrifugal force.

N+$\bf{ m\omega^{2}r}$ = mg

(see the attachment! )

☆Solution:

》for the condition of not spilling, the mg must be balanced.

》and that is balanced by centrifugal force, considering the minimum case of rotation(where water is in the verge of spill,but does not)

》So taking Normal as almost negligible.

$\rightarrow{\sf{ m\omega^{2}r = mg}}$

$\rightarrow{\sf{ \omega^{2}r = g}}$

$\rightarrow{\sf{ \omega^{2}= \dfrac{g}{r}}}$

$\rightarrow{\sf{ \omega= \sqrt{\dfrac{10}{5}}}}$

$\rightarrow{\bf{\omega= 1.4rad/sec}}$

》rad=1/2$\pi$ revolution.

》rad/sec=60/2$\pi$ revolution per min

》rad/sec = 30/$\pi$ rev/min

$\bf{ \omega = 1.4× \dfrac{30}{\pi}}$

$\bf{ \omega = 13.4 rev\;per\;min}$

☆considering revolutions are in one minute, almost

no. of revolutions(n) = 13

☆☆13 revolutions are necessary for water to not spill.

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♧hope it helps!♧

Answer 2

Please see the attachment