Question

A bucket tied at the end of 1.6 m long string is whirled in a vertical circle, when the bucket is at the highest point, the minimum speed for the bucket not to spill water from it is (take g = 10 ms)

Answer 1

Concept:

• In circular motion,bucket will perform a circular path through a fixed distance known as radius around a fixed point.
• centrifugal force is experienced by the bucket outwards due to the tension in string.

Solution:

For not to spill water from bucket at highest point :

$\boxed{\bf centrifugal\: force \geqslant mg}$

$\to\rm \dfrac{mv^2}{r} \geqslant mg$

$\to\rm \dfrac{v^2}{r} \geqslant g$

$\to\rm v^2 \geqslant rg$

$\to \bf v \geqslant \sqrt{rg}$

$\to \rm v \geqslant \sqrt{1.6\times 10}$

$\to \rm v \geqslant \sqrt{16}$

$\to \bf v \geqslant 4 \: ms^{-1}$

★ minimum speed for the bucket not to spill water is 4m/s.

Answer 2

Answer:

A bucket tied at the end of 1.6 m long string is whirled in a vertical circle, when the bucket is at the highest point, the minimum speed for the bucket not to spill water from it is (take g = 10 ms)

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A bucket tied at the end of 1.6 m long string is whirled in a vertical circle .

The minimum speed for the bucket not to split water from it is = 10 m/s .

[ Let's find out Critical velocity at height point ]

=> √gR

Value of g = 10 and value of R = 1.6

Putting the value :

= √10×1 .6

=√16

= 4 m/s

So the minimum speed for the bucket = 4 m/sec .

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