Question 269: Work, Energy and Power & Circular Motion

Question

A bucket containing water is revolved in a vertical circle of radius $$r$$. To prevent the water from falling down, the minimum frequency of revolution required is ($$\mathrm{g}=$$ acceleration due to gravity)

Accepted Answer

To answer this question, we need to consider the forces in action when the bucket is at the topmost point in its circular path. At that point, the centripetal force required to keep the water moving in a circular path must be greater than or equal to the gravitational force acting on the water, to prevent it from falling out of the bucket.
The centripetal force ($$ F_c $$) can be described by the following equation, where $$ m $$ is the mass of water, $$ v $$ is the linear velocity of the bucket, and $$ r $$ is the radius of the circle:
$$ F_c = \frac{m v^2}{r} $$
At the minimum velocity needed to keep the water in the bucket, this centripetal force is provided entirely by the weight of the water, which is $$ mg $$, where $$ g $$ is the acceleration due to gravity. Therefore, we can set $$ F_c = mg $$ and solve for the velocity:
$$ mg = \frac{m v^2}{r} $$
dividing both sides by $$ m $$ and then multiplying by $$ r $$ gives us:
$$ v^2 = rg $$
Now, velocity can also be related to the frequency of revolution ($$ f $$) and the circumference of the circle ($$ C $$) using the relation:
$$ v = f 	imes C $$
The circumference of the circle is given by:
$$ C = 2 \pi r $$
Let's substitute this into the velocity equation and solve for frequency:
$$ \sqrt{rg} = f 	imes 2 \pi r $$
dividing both sides by $$ 2 \pi r $$ gives us:
$$ \frac{\sqrt{rg}}{2 \pi r} = f $$
To isolate $$ f $$, since $$ r $$ is in a square root in the numerator and is not in a square root in the denominator, we can simplify:
$$ f = \frac{1}{2 \pi} \sqrt{\frac{g}{r}} $$
So the correct answer to the minimum frequency required to prevent the water from falling out of the bucket is Option C:
$$ \frac{1}{2 \pi} \sqrt{\frac{g}{r}} $$

Answer

$$2 \pi \sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$$

Answer

$$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{r}}{\mathrm{g}}}$$

Answer

$$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}$$

Answer

$$2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}$$

Work, Energy and Power & Circular Motion

A bucket containing water is revolved in a vertical circle of radius $r. To prevent the water from falling down, the minimum frequency of revolution required is (\mathrm{g}=$ acceleration due to gravity)

Choose the correct answer: