Answer to Question #158244 in Classical Mechanics for Jay mishra

Question

Four point masses of mass each of 1 kg are connected by a string which forms a square of diagonal 0.707 m . If square is placed on rotating table rotating with 5 rotation per second then what is the tension in string ?

Answer 1

Answer

Given diagonal =0.707 meter

Lets take length of square is a meter diagonal will be

$\sqrt{2}$ a meter

​

$\sqrt{2}a$ =0.707 meter

a=0.5 meter

Rotation is w=5 rps

=5×2π=10πrad/s

We all know that

Radius for revolution = half of diagonal

$=\frac{a}{\sqrt{2}}$

So we can write

$2T\cos 45^°=mw^2r$

Putting value of m, w, r then the tension

T=246.8N