Question

One end of a 1.0-m string is fixed, the other end is attached to a 2.0-kg stone. The stone swings in a vertical circle, passing the top point at 4.0 m/s. The tension force of the string (in newtons) at this point is about:

Answer 1

Given:

One end of a 1.0-m string is fixed, the other end is attached to a 2.0-kg stone. The stone swings in a vertical circle, passing the top point at 4.0 m/s.

To find:

Tension force at this point ?

Calculation:

According to FBD of the body in the topmost point in the vertical circle , we can say:

• Tension is directed downwards
• Weight is also directed downwards

$\therefore \: T + mg = \dfrac{m {v}^{2} }{r}$

$\implies\: T +2g= \dfrac{2 \times {(4)}^{2} }{1}$

$\implies\: T +2(10)= \dfrac{2 \times {(4)}^{2} }{1}$

$\implies\: T +20= \dfrac{2 \times {(4)}^{2} }{1}$

$\implies\: T +20= 32$

$\implies\: T = 32 - 20$

$\implies\: T = 12 \: N$

So, tension at that point is 12 N.