Question

A 0.150 kg ball on the end of a 1.10 m long cord (negligible mass)is swung in a vertical circle..

(a) what is the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle?
calculate the tension in the cord at the bottom of the arc assuming the ball is moving at twice the speed of part (a) ...?

Answer 1

For any body to move in a circle it requires the centripetal force (mv^2)/r. In this case a ball is moving in a vertical circle swung by a mass less cord. At the top of its arc if we draw its free body diagram and equate the forces in radial direction to the centripetal force we get it as T +mg =(mv^2)/r T is tension in cord m is mass of ball r is length of cord (radius of the vertical circle) To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s In the second case the speed of ball at top = (2*3.285) =6.57 m/s Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom we get velocity at bottom as 9.3m/s. Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force T-mg=(mv^2)/r We get tension in cord T=13.27 N

Answer 2

`v = 3.28 m/s`

Step-by-step explanation:

At the top position of the arc if ball continue in its circular path then we can say that tension force must be zero at that position for the minimum value of the speed.

So we will have

`T + mg = (mv^2)/(R)`

so if minimum speed is required to complete the circle T = 0

`0 + mg = (mv^2)/(R)`

so we will have

`v = √(Rg)`

again we will have

`v = √(1.10* 9.81)`

`v = 3.28 m/s`