) A man holds a pail of water by the handle and whirls it around a vertical circle of radius 0.86 m at constant speed. What is the minimum speed that the pail must have at the top of its circular motion if the water is not to spill out of the upside-down pail?

centripetalforce=mg

m*v^2/r=mg
v= sqrt (rg)

Well, if the pail is upside-down, I guess you could say the water is feeling a little "upended." To keep it contained, we need to find the minimum speed at the top of the circular motion.

To avoid any spills, we need to consider the force of gravity working against the centrifugal force. If the speed is too slow, the water will just happily say "see ya" and spill out.

Applying the concept of circular motion dynamics, we know that the centripetal force is equal to the weight of the water in the pail. So, we have the equation:

mv^2/r = mg

Where:
m = mass of the water
v = velocity of the pail at the top
r = radius of the circular motion
g = acceleration due to gravity

We need to solve for v, so let's rearrange the equation:

v^2 = rg

Finally, taking the square root of both sides, we get:

v = sqrt(rg)

Plugging in the values, with r = 0.86 m and g = 9.8 m/s^2 (approximately), we can calculate the minimum speed required at the top:

v = sqrt(0.86 * 9.8) m/s

Don't worry, I'll do the math. Just give me a second... There we go!

The minimum speed the pail must have at the top is approximately 8.53 m/s. So, if the pail is "feeling the need for speed" at that rate, the water will stay safely inside. Happy spinning!

To find the minimum speed that the pail must have at the top of its circular motion, we can use the concept of centripetal acceleration.

First, let's consider the forces acting on the water in the pail. At the top of the circular motion, the net force on the water must provide the necessary centripetal force to keep the water from spilling out of the pail.

The force acting on the water at the top is the weight of the water, which is given by:

Weight = mass of water x acceleration due to gravity

Next, we can use Newton's second law to relate the net force and the centripetal force:

Net force = mass of water x centripetal acceleration

Since the net force is equal to the weight, we can equate the expressions for weight and net force:

mass of water x acceleration due to gravity = mass of water x centripetal acceleration

The mass of the water cancels out, leaving us with:

acceleration due to gravity = centripetal acceleration

Now, let's calculate the value of the centripetal acceleration at the top of the motion. The centripetal acceleration can be calculated using the equation:

centripetal acceleration = (velocity^2) / radius

where velocity is the speed of the pail at the top, and radius is the radius of the circular motion.

Plugging in the values:

acceleration due to gravity = (velocity^2) / radius

Rearrange the equation to solve for velocity:

velocity^2 = acceleration due to gravity x radius

velocity = sqrt(acceleration due to gravity x radius)

Now, let's calculate the minimum speed by substituting the given values:

acceleration due to gravity = 9.8 m/s^2 (approximate value)
radius = 0.86 m

velocity = sqrt(9.8 m/s^2 x 0.86 m)

velocity = sqrt(8.428 m^2/s^2)

velocity ≈ 2.9 m/s

Therefore, the minimum speed that the pail must have at the top of its circular motion is approximately 2.9 m/s.

To determine the minimum speed that the pail must have at the top of its circular motion, we need to consider the forces acting on the water inside the pail. At the top of the circular motion, the force of gravity acting on the water is directed towards the center of the circular path, while the inertia of the water tries to make it move tangentially away from the circular path.

For the water to remain inside the pail, the force of gravity must be greater than or equal to the centrifugal force acting on the water. The centrifugal force is given by the equation:

Fc = m * v^2 / r

Where:
Fc is the centrifugal force
m is the mass of the water
v is the speed of the pail at the top of the circular motion
r is the radius of the circular path

Since we want the water to remain in the pail, the centrifugal force must not exceed the force of gravity. Therefore, we can equate the centrifugal force to the force of gravity:

Fc = m * v^2 / r = m * g

Where:
g is the acceleration due to gravity (approximately 9.8 m/s^2)

By rearranging the equation, we can solve for the minimum speed (v) required at the top of the circular motion:

v^2 = r * g

v = sqrt(r * g)

Substituting the given radius (r = 0.86 m) and acceleration due to gravity (g = 9.8 m/s^2) into the equation:

v = sqrt(0.86 * 9.8) ≈ 2.97 m/s

Therefore, the minimum speed that the pail must have at the top of its circular motion is approximately 2.97 m/s in order to prevent the water from spilling out.