Circular Motion

Question:

An object of mass 6.0 kg is whirled round a vertical circle of radius 2.0m with a speed of 8.0m/s.
(a) calculate the maximum and minimum tension in the string connecting the object to the centre of the circle.
(b) if the string breaks when the tension in it exceeds 360N, calculate the maximum speed of rotation in m/s, and state where the object will be when it breaks.

Answer:

Mass of object = 6.0 kg, radius of circle = 2.0 m speed = 8.0 m/s
The difficulty here is of course because the circle is vertical. In this case the tension in the string will vary as the object goes round being greatest at the bottom. The centripetal force is always the same but part of this (and a constant part – the weight of the object) is supplied by the gravitation al pull of the planet.

(a) The minimum tension is at the top.
Centripetal force = constant (constant speed) = mv²/r = 192 N.
Tension in the string = centripetal force – weight = 192 - 6.0g = 192 - 58.86 N
= 133.14 = 133 N
The maximum tension is at the bottom.
Centripetal force = constant (constant speed) = mv²/r = 192 N.
Tension in the string = centripetal force – weight = 192 + 6.0g = 192 + 58.86 N = 250.86 = 251 N
(I have assumed g = 9.81 m/s²).

(b) The object will be at the bottom where the gravitational attraction (weight) is no help in keeping the object moving in a circle.
Centripetal force = constant (constant speed) = mv²/r.
Tension in the string = 360 = centripetal force – weight = centripetal force
Therefore: Centripetal force = mv²/r = 360 + 58.86 = 418.86 N
For the values given above for the mass of the object and the radius of the "orbit" this gives a value of 11.8 m/s for the velocity.