Physics, 19.09.2020 01:01 tangia

Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.82 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 18.5 cm, what minimum speed is required?

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Answer from: sequoyaburke

Approximately $3.1\; \rm m \cdot s^{-1}$ .

Explanation:

The content of this pail is in a centripetal motion because its path forms part of a vertical circle. Let denote the mass of the contents of this pail, let denote the (linear) velocity of the content, and let denote the radius of this circle. The net force on the contents of this pail will thus be:

$\displaystyle F(\text{net}) = \frac{m\, v^2}{r}$ towards the center of the circle.

Assume that there is no friction between the content and walls of the pail. The only two possible forces on the contents pail towards the center would be:

The downwards gravitational pull from the earth, Normal force between walls of the pail and the contents (except at the top and bottom of the circle,) andIf the rotation is fast enough, the normal force from the bottom of the pail, which also points downwards.

Note that at the top of the circle, both the gravitational pull and the normal force from the bottom point towards the center of the circle. On the other hand, the normal force from the walls of the pail would be perpendicular to the line towards the center of the circle. At that point in the circle, there's no upward force to support the content of the pail. The uniform rotation will be sufficiently fast if it could allow the content to stay in the pail at the top of the circle.

Let denote the gravitational field strength at the top of this circle. The size of the gravitational pull on the content would be $m\cdot g$ . Let $F(\text{normal})$ denote the normal force from the bottom of the pail on the contents. The sum of these two forces should be equal to the vertical net force on the contents of this pail. That is:

$F(\text{net}) = m\cdot g + F(\text{normal})$ .

From the centripetal motion of the content:

$\displaystyle \frac{m\, v^2}{r} = m\cdot g + F(\text{normal})$ .

Rearrange to obtain an expression for the normal force:

$\displaystyle F(\text{normal}) = \frac{m\, v^2}{r} - m\cdot g$ .

Note, that the normal force the bottom of the pail exerts on the contents should be greater than or equal to zero. While the pail is at the top of the circle, the normal force from the bottom of the pail cannot pull the contents upwards. Hence:

$\displaystyle \frac{m\, v^2}{r} - m\cdot g = F(\text{normal}) \ge 0$ .

$\displaystyle \frac{m\, v^2}{r} - m\cdot g \ge 0$ .

Rearrange and simplify to obtain:

$\displaystyle \frac{v^2}{r} - g \ge 0$ .

$v^2 \ge g\cdot r$ .

$v \ge \sqrt{g \cdot r}$ .

In other words, if the gravitational field strength is and the radius of the circle is , the minimum linear velocity required to keep the content in the pail at the top of the circle is $\sqrt{g \cdot r}$ .

If $g = 9.81\; \rm N \cdot kg^{-1} = 9.81\; \rm m \cdot s^{-2}$ and $r = 0.82 \; \rm m + 0.185\; \rm m \approx 1.005\; \rm m$ , then the minimum value of would be approximately:

$\sqrt{9.81 \; \rm m \cdot s^{-1} \times 1.005\; \rm m} \approx 3.1\; \rm m \cdot s^{-1}$ .

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

He can play two games. 13/4=3.253.25*2=6.5

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Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its cont...