Uniform Circular Motion of a ball

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SUMMARY

The discussion focuses on calculating the tension in a string when a ball of mass 0.300 kg is revolved in a vertical circle with a radius of 65 cm at a speed of 4.25 m/s. The key equations used include the centripetal acceleration formula, a_r = v^2/R, and the net force equation, Fnet = Fg + Ft, where Fg is the force of gravity and Ft is the tension in the string. The tension is highest at the bottom of the path and lowest at the top, with the net force determined by the centripetal acceleration and gravitational force acting on the ball.

PREREQUISITES
  • Understanding of uniform circular motion principles
  • Familiarity with Newton's second law (F = ma)
  • Knowledge of centripetal acceleration (a_r = v^2/R)
  • Basic concepts of forces, including tension and gravity
NEXT STEPS
  • Calculate tension in the string at the top and bottom of the circular path using the derived equations
  • Explore the effects of varying mass and speed on tension in uniform circular motion
  • Study the relationship between radius and centripetal force in circular motion
  • Investigate real-world applications of uniform circular motion in engineering and physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of tension and forces in action.

jewel203
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i'm having trouble in solving this problem. I've tried several ways to solve this problem but still have not come up with the correct solution. the question is:

A ball on the end of a string is revolved at a uniform rate in a vertical circle of radius 65 cm. If its speed is 4.25 m/s and its mass is .300 kg, calculate the tension in the string when the ball is (a) at the top of its path, and (b) at the bottom of its path.

i tried using the equation: (EF)=ma...but i think you use that later on?
there's another equation that i think i might have to use to solve this problem, however I am not sure on how to find v1 and v2. this equation is:
FT1 + mg = m * (v2,1/r).
 
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jewel203 said:
i tried using the equation: (EF)=ma...but i think you use that later on?

ALLL problems in mechanics find their answer in this equation, and this one is no exception.

The one equation that you will need besides F=ma is the expression of the acceleration of an object in uniform circular motion:

[tex]a_r = v^2/R[/tex]

You are basically asked to find what must the tension in the string be at these two points so that the sum of the forces on the ball always be equal to [itex]ma_r[/itex].


P.S. In case you didn't knew, writting "EF" doesn't make much sense. I think you meant [itex]\Sigma \vec{F}[/itex]? This sign before F is the greek letter "capital sigma" and it usually stands in mathematical notation for "sum". So [itex]\Sigma \vec{F} = m \vec{a}[/itex] means "the sum of the forces (also called the 'net force') acting on a particle is equal to its mass times its acceleration".
 
jewel203 said:
i tried using the equation: (EF)=ma...
What's the E for?

There are two forces acting on the ball: The tension of the string, which provides the centripetal force, and gravity. For an object in uniform circular motion, the centripetal acceleration is equal to [itex]v^2/r[/itex].

We know that the net force acting on the ball is the sum of the two individual forces,
Fnet = Fg + Ft
Where Fg is the force of gravity, and Ft is the tension in the string.

So the tension must be the difference between the net force and the force of gravity,
Fnet - Fg = Ft

In order to evaluate this expression for Ft, we need to know what Fnet and Fg are. The force of gravity is easy, it's just the product of the mass of the object and the acceleration due ot gravity. What about Fnet? Well, since we know the ball is moving in a uniform circle, we know the acceleration is [itex]v^2/r[/itex], and the problem gives us the mass. From F = ma, we just multiply those two things together to find the net force.

You should decide which direction you want to make positive, and which direction you want to make negative. It's probably most intuitive to say that "up" is positive. If so, then when the ball is at the top of its swing, it has a net acceleration of [itex]-v^2/r[/itex], and when it is at the bottom of its swing, it has a net acceleration of [itex]+v^2/r[/itex]. Don't forget that this would mean Fg is a negative quantity, as well. The tension in the string should be greatest when the ball is at the bottom of its path, and least when it is at the top.

Does this answer your question?
 

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