Name: Conservation of Mechanical Energy
Uploaded: 2019-07-19
Duration: 5 min 12 s
Description: When a body falls from a height towards the ground, its potential energy decreases but not the total mechanical energy. Let us understand this using data.

Question

A test tube of mass $10 grams$ closed with a cork of mass $1 gram$ contains some ether. When the test tube is heated the cork flies out under the pressure of ether gas. The test tube is suspended horizontally by a weightless rigid bar of length $5 cm$ . What is the minimum velocity with which the cork should fly out the tube so that the test tube describing a full vertical circle about the point O? Neglect the mass of the ether.

Accepted Answer

We know that the minimum velocity for one complete in a vertical circle, the minimum velocity at the lowest point:

$v = \sqrt{5 g R}$

Given:

$g = 10 m / s^{2}$

The radius of the circle of revolution will be equal to the length of the rigid bar. Hence:

The radius of the circle of revolution $(R) = 5 cm = 0.05 m$

Mass of the test tube $(M) = 10 g$

The minimum velocity of the test tube at the lowest point of the vertical circle, so that it can complete full vertical circle is given by:

$(v) = \sqrt{5 g R}$

$(v) = \sqrt{5 \times 10 \times 0.05} = \sqrt{2.5} m / s$

The test tube will get this velocity as the cork is ejected from the tube. Using conservation of linear momentum for the test tube cork system. Let the velocity of cork be $u$ and its mass be $m$ . Then:

$m u = - M v$

$\Rightarrow u = \frac{- M v}{m} = - \frac{10}{1} \times \sqrt{2.5} = - 5 \sqrt{10} m / s$

Hence, the cork must eject out with this speed in the direction opposite to the direction of motion of the test tube.

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