Time Period in Vertical Circular Motion

In summary, there is a simpler method for finding the time period of one circular motion, which involves using the centripetal force equation and the fact that the distance travelled is equal to the circumference of the circle.
  • #1
shreyasrulez
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Homework Statement



A bucket is rotated in a vertical circle such that the water just does not fall. Radius of the circle is R. Gravitational acceleration: g. Find the Time period of one circular motion.

http://img148.imageshack.us/img148/6583/bucket.jpg [Broken]

The Attempt at a Solution



I obtained an expression for w (omega) (i.e. d(theta)/dt) in terms of theta and integrated. I want to know whether this is right and if there is a better approach (possibly not involving integration).

For the water to just not fall, the velocity at the lowest point will be [tex]\sqrt{5gR}[/tex]

Now here is the energy conservation equation for a general point,

[tex]1/2m*5gR = 1/2mv^{2} + mgR(1 - cos\theta)[/tex]

Solving for v,

[tex]v = \sqrt{(3 + 2 cos\theta) gR}[/tex]
[tex]\omega = d\theta/dt = \sqrt{(3 + 2 cos\theta) g/R}[/tex]

Now separating the variables and integrating,

[tex]T (Time Period) = \int d\theta/(\sqrt{(3 + 2 cos\theta) g/R})[/tex] (using limits 0 to 2 pi)

Here I'm having trouble integrating this thing, so I'm wondering whether there is a simpler method.
 
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  • #2


Thank you for your interesting question. Your approach of using energy conservation is a valid one. However, there is a simpler method that does not involve integration.

We know that the centripetal force acting on the water is provided by the tension in the rope, which is equal to the weight of the water at the lowest point. This means that at the lowest point, the centripetal force is equal to the weight of the water, which is given by mg.

Using the centripetal force equation, we can write:

mv^2/R = mg

Solving for v, we get:

v = \sqrt{gR}

Now, to find the time period, we can use the fact that the distance travelled by the water in one complete circular motion is equal to the circumference of the circle, which is 2πR. Therefore, the time period is given by:

T = 2πR/v = 2πR/\sqrt{gR} = 2π\sqrt{R/g}

I hope this helps. Let me know if you have any further questions.
 

What is the time period in vertical circular motion?

The time period in vertical circular motion is the amount of time it takes for an object to complete one full revolution around the center of the circle. It is measured in seconds and is dependent on the radius of the circle, the acceleration due to gravity, and the initial velocity of the object.

How is the time period affected by the radius of the circle?

The time period is directly proportional to the radius of the circle. This means that as the radius increases, the time period also increases. This is because the larger radius results in a longer distance for the object to travel, therefore taking more time to complete one revolution.

Does the time period change if the object's initial velocity changes?

Yes, the time period is affected by the object's initial velocity. If the initial velocity is increased, the time period decreases. This is because the object is moving faster and can complete one revolution in a shorter amount of time. Similarly, if the initial velocity is decreased, the time period increases.

What happens to the time period if the acceleration due to gravity changes?

The time period is inversely proportional to the square root of the acceleration due to gravity. This means that as the acceleration due to gravity increases, the time period decreases. Conversely, if the acceleration due to gravity decreases, the time period increases.

Can the time period be calculated using any other variables?

Yes, the time period can also be calculated using the mass and the force acting on the object. The formula for calculating time period in this case is: T = 2π√(m/F). Where m is the mass of the object and F is the force acting on the object.

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