angular velocity

[shelly89]

1)A light bucket contains a ball of mass m. The bucket is attached to a rigid rod and swung in a vertical circle at a constant angular velocity. The distance from the pivot point to the bottom of the bucket is a.

What is the minimum angular velocity of the bucket such that the ball does not fall out of the bucket?

am really not understand how to picture this scenario, or even how to start.

but the ball has a weight W = -mgK

is there a normal reaction force in the z direction? and than theres a centripetal force, pushing the ball to stay at the bottom of the bucket?

am not sure how to put this into equations.

any help appreciated.

 

[wjm11]

1)A light bucket contains a ball of mass m. The bucket is attached to a rigid rod and swung in a vertical circle at a constant angular velocity. The distance from the pivot point to the bottom of the bucket is a.

What is the minimum angular velocity of the bucket such that the ball does not fall out of the bucket?

am really not understand how to picture this scenario, or even how to start.

but the ball has a weight W = -mgK

is there a normal reaction force in the z direction? and than theres a centripetal force, pushing the ball to stay at the bottom of the bucket?

am not sure how to put this into equations.

any help appreciated.

Understanding the problem: Suppose you had a bucket of water that you start to swing in an arc, back and forth. Finally, you swing the bucket in a big circle above your head and back near the ground again. If you swing the bucket in that circle fast enough, the water will stay in the bucket even when it's above your head.

This problem is the same thing, except the water is replaced by a ball in the bucket.

If you continue to swing this bucket in a circle repeatedly, you will notice something: The pull against your arm is very light when the bucket is at its highest point, and the pull is very strong when the bucket is at its lowest point.

This is due to the way two forces are combining at different points in the arc. The two forces are gravity (naturally, since it's always present) and centripetal force (a result of the circular motion of the bucket). Centripetal acceleration is a = v^2/r, and centripetal force is F = ma = mv^2/r.

If you swing the bucket at just the right speed, the bucket will seem weightless at the top of the arc, and it will seem to be double its weight at the bottom of the arc. This is the minimum speed required to keep the ball (or water) in the bucket.

To see this explained in greater detail, visit: http://www.physicsclassroom.com/class/circles/Lesson-1/Mathematics-of-Circular-Motion

 

[shelly89]

Understanding the problem: Suppose you had a bucket of water that you start to swing in an arc, back and forth. Finally, you swing the bucket in a big circle above your head and back near the ground again. If you swing the bucket in that circle fast enough, the water will stay in the bucket even when it's above your head.

This problem is the same thing, except the water is replaced by a ball in the bucket.

If you continue to swing this bucket in a circle repeatedly, you will notice something: The pull against your arm is very light when the bucket is at its highest point, and the pull is very strong when the bucket is at its lowest point.

This is due to the way two forces are combining at different points in the arc. The two forces are gravity (naturally, since it's always present) and centripetal force (a result of the circular motion of the bucket). Centripetal acceleration is a = v^2/r, and centripetal force is F = ma = mv^2/r.

If you swing the bucket at just the right speed, the bucket will seem weightless at the top of the arc, and it will seem to be double its weight at the bottom of the arc. This is the minimum speed required to keep the ball (or water) in the bucket.

To see this explained in greater detail, visit: http://www.physicsclassroom.com/class/circles/Lesson-1/Mathematics-of-Circular-Motion

I understand the situation, but dont quire understand how to put this into formula.

how do

 

[wjm11]

I understand the situation, but dont quire understand how to put this into formula.

how do

The short answer: You need to calculate the velocity that will result in centripetal acceleration being equal to the acceleration due to gravity, g = 9.81 m/s^2.

So, centripetal acceleration a = g = v^2/r

The reason: An object will always travel in a straight line unless acted upon by an outside force. For an object to travel in a circle, the net "outside force" must always act in the direction of the center of the circle. This results in an acceleration toward the center of the circle.

When we are swinging a bucket in a circular plane, there are always two separate force components to consider: the force of gravity (which is always downward) and the force applied by our arm/string/rod/whatever (which is always toward the center of rotation).

When the bucket is at the bottom of its arc, we must apply enough force to both keep the bucket in its circular orbit as well as offset its static weight.

When the bucket is at the top of its arc, we can let gravity apply the necessary (instantaneous) force to keep the bucket in its circular orbit, and we don't have to add any force ourselves. That is why we set gravity equal to the centripetal acceleration to solve this problem.

 

[shelly89]

The short answer: You need to calculate the velocity that will result in centripetal acceleration being equal to the acceleration due to gravity, g = 9.81 m/s^2.

So, centripetal acceleration a = g = v^2/r

The reason: An object will always travel in a straight line unless acted upon by an outside force. For an object to travel in a circle, the net "outside force" must always act in the direction of the center of the circle. This results in an acceleration toward the center of the circle.

When we are swinging a bucket in a circular plane, there are always two separate force components to consider: the force of gravity (which is always downward) and the force applied by our arm/string/rod/whatever (which is always toward the center of rotation).

When the bucket is at the bottom of its arc, we must apply enough force to both keep the bucket in its circular orbit as well as offset its static weight.

When the bucket is at the top of its arc, we can let gravity apply the necessary (instantaneous) force to keep the bucket in its circular orbit, and we don't have to add any force ourselves. That is why we set gravity equal to the centripetal acceleration to solve this problem.

if your solving for v, than isnt the answer \(\displaystyle \sqrt{ga} = v \) ?

my book says the correct answer is

\(\displaystyle \dot{\phi} > \sqrt{\frac{g}{a}} \)

I dont understand how they got this? can anyone explain? ( I know we are using newtons second law in polar coordinates )

thanks

 

[wjm11]

if your solving for v, than isnt the answer \(\displaystyle \sqrt{ga} = v \) ?

my book says the correct answer is

\(\displaystyle \dot{\phi} > \sqrt{\frac{g}{a}} \)

I dont understand how they got this? can anyone explain? ( I know we are using newtons second law in polar coordinates )

thanks

I have given you the lowest instantaneous velocity solution, NOT the angular velocity. You need to convert it.

If you need help on angular velocity, check these sites:

http://en.wikipedia.org/wiki/Angular_velocity

https://www.khanacademy.org/science...lationship-between-angular-velocity-and-speed

http://people.wku.edu/david.neal/117/Unit2/AngVel.pdf