A stone tied to the end of a string 80 cm long is whirled in a horizon

1.	A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolution is 25 s, what is the magnitude and direction of acceleration of the stone?
Answer» Here, r = 80 cm = 0.80 m, u = 14 rev/25 s As centripetal acceleration = 4π²v²r = 4 x (3.14) 2 x (14/25)² x 0.80 = 9.9 ms^-2 At every point the acceleration is along the radius and towards the centre.

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolution is 25 s, what is the magnitude and direction of acceleration of the stone?

Answer»

Here, r = 80 cm = 0.80 m,

u = 14 rev/25 s

As centripetal acceleration = 4π²v²r

= 4 x (3.14) 2 x (14/25)² x 0.80 = 9.9 ms^-2

At every point the acceleration is along the radius and towards the centre.