Answer to Question #146068 in Physics for Khan Asf

Question

A small ball of mass O, 5 kg hangs on a non-stretchable cord. The cord is placed horizontally and then the ball is released. Find the tension force of the cord when it forms angle 30° with the vertical. The answer should be expressed in newtons [N]. Assume the free fall acceleration to be 10 m/s^2.

Answer 1

Find the speed of the ball when the cord (length $l$) makes a 30° angle:

$$v=\sqrt{2gh}=\sqrt{2gl\text{ sin}30°}.$$

Find the centripetal acceleration at this speed:

$$a=\frac{v^2}{l}=2g\text{ sin}30°.$$

The acceleration creates a force of

$$F=ma=2mg\text{ sin}30°.$$

The force of gravity that acts on the cord at this angle is

$$F_g=mg\text{ cos}60°.$$

So, the tension is

$$T=F+F_g=\\=mg[\text{cos}(90°-30­°)+2\text{sin30°}]=7.5\text{ N}.$$