A ball on the end of a string is cleverly revolved at a uniform rate in a vertic

Question

A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 83.0cm, as shown in figure above

If its speed is 4.12m/s and its mass is 0.271kg, calculate the magnitude of the tension in the string when the ball is at the top of its path.

A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 83.0cm, as shown in figure above If its speed is 4.12m/s and its mass is 0.271kg, calculate the magnitude of the tension in the string when the ball is at the top of its path.

Explanation / Answer

When it is at top of its path,

the net force, Fnet = mg + T

where T = tension in string

m = mass

g =9.8 m/s2

r= radius of path = 83 cm = 0.83 m

and they are in same direction,

so, Fnet = mg + T

and this Fnet must provide the necessary centripetal force

so, Fnet = mg + T = mv^2/r

So, T = mv^2/r-mg = 0.271*4.12^2/0.83 - 0.271*9.8 = 2.89 N <-------------answer

b)

at bottom, Fnet = T - mg = mv'^2/r

so, T = mg + mv'^2/r <------here v' = velocity at bottom of the circle

By conservation of energy,

K.E + P.E at the top = K.E at the bottom

So, 0.5*mv^2+mg(2r) = 0.5*mv'^2 <-----here as from the top to the bottom, the ball descends by = 2r heoght, so, P.E = mgh = mg(2r)

So, v'^2 = (v^2+4gr)

so, T = mg+mv^2/r = mg +m*(v^2+4gr)/r

= 0.271*9.8+0.271*(4.12^2+4*9.8*0.83)/0.83

= 18.82 N <----------------answer

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