Question

A pail of water of mass 2000 gm is rotated in a vertical circle of radius 1.00 m. What is the pail’s minimum speed at the top of the circle if no water is to spill out and the tension exerted by the string is 25 N?

Answer 1

The pail’s minimum speed at the top of the vertical circle is 4.72 m/s.

The given parameters:

• Mass of the water, m = 2000 g = 2 kg
• Radius of the circle, r = 1 m
• Tension on the string, T = 25 N

The net force on the pail of water at the top of the vertical circle is calculated as follows;

$T = ma - mg\\\\T = \frac{mv^2}{r} - mg\\\\T + mg = \frac{mv^2}{r} \\\\mv^2 = r(T + mg)\\\\v^2 = \frac{r(T + mg)}{m} \\\\v= \sqrt{\frac{r(T + mg)}{m}} \\\\v = \sqrt{\frac{1 (25\ + \ 2 \times 9.8)}{2}} \\\\v = 4.72 \ m/s$

Thus, the pail’s minimum speed at the top of the vertical circle is 4.72 m/s.

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Answer 2

The pail’s minimum speed at the top of the circle if no water is to spill out is; v = 4.722 m/s

We are given;

Mass of the pail of water; m = 2000 g = 2kg

Radius of the circle; r = 1 m

Tension exerted by string; T = 25 N

To calculate the net force on the pail of water at the top of the vertical circle, let us take equilibrium of forces to give;

W - T = ma

Where;

W is weight = mg

T is tension on the string

Now, in circular motion, we know that;

Acceleration is; a = v²/r

Thus;

mg - T = m(v²/r)

Making v the subject of the formula gives us;

v = √[(r(T + mg)/m)

Plugging in the relevant values gives;

v = √[(1(25 + (2 * 9.8))/2)

v = 4.722 m/s

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