Centripital Motion: Find Tension & Speed of Bead

  • Thread starter chopramon
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In summary: The bead is moving at a speed of v = F/ma. The centripital force (F) is equal to the tension in the string (T) and the bead is moving in a straight line. Therefore, F = T.
  • #1
chopramon
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1. A 100g bead is free to slide along a 80cm piece of string ABC. Points A and B of the string are attached to a vertical pole 40 cm apart. When the pole is rotated AB becomes horizontal.

I have attached a diagram of the question.

Find the tension in the string?
Find the speed of the ball at point B?

Relevant Equations
F = mv^2/r
a = v^2/r
F = ma

Attempt:
I know that the centripital force acting on the object must be equal to the Tension acting in the horizontal direction. Also the radius of the string when the bead is at point B is a function of the angle. I also believe that the vertical force acting on the string must be equal to the downward force of gravity. Therefore so far I have:

1. radius = 40/tan0
2. a = v^2/(40/tan0) (this is when subbing in for r the above equation)
3. Ty = mg = 0.98

I am not sure where to go from here, or even if I am starting off right. Any help would be appreciated.

Thanks
 

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  • #2
chopramon said:
1. A 100g bead is free to slide along a 80cm piece of string ABC. Points A and B of the string are attached to a vertical pole 40 cm apart. When the pole is rotated AB becomes horizontal.

I have attached a diagram of the question. Does anybody know how to put the picture of the question directly on this post?

Find the tension in the string?
Find the speed of the ball at point B?

Relevant Equations
F = mv^2/r
a = v^2/r
F = ma

Attempt:
I know that the centripital force acting on the object must be equal to the Tension acting in the horizontal direction. Also the radius of the string when the bead is at point B is a function of the angle. I also believe that the vertical force acting on the string must be equal to the downward force of gravity. Therefore so far I have:

1. radius = 40/tan0
2. a = v^2/(40/tan0) (this is when subbing in for r the above equation)
3. Ty = mg = 0.98

I am not sure where to go from here, or even if I am starting off right. Any help would be appreciated.

Thanks
 
  • #3
You have the info needed to find the sides and angles of the right triangle.

Apply Newton's 2nd law to the horizontal components.
 

1. What is centripetal motion?

Centripetal motion is the movement of an object along a circular path with a constant speed.

2. How is tension related to centripetal motion?

Tension is the force that keeps an object moving in a circular path. In centripetal motion, tension is directed towards the center of the circle and is responsible for changing the direction of the object's velocity.

3. How do you calculate the tension in a centripetal motion problem?

The tension can be calculated using the centripetal force formula, which is T = (mv^2)/r, where T is tension, m is mass, v is velocity, and r is the radius of the circular path.

4. How do you find the speed of an object in centripetal motion?

The speed can be calculated using the formula v = (2πr)/T, where v is velocity, r is the radius, and T is the time taken for one complete revolution.

5. Can the tension in a centripetal motion problem ever be greater than the weight of the object?

Yes, the tension can be greater than the weight of the object in centripetal motion if the speed of the object is high enough. This is because the tension must provide enough centripetal force to keep the object moving in a circular path, which increases with higher speeds.

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