A stone tied to the end of a string 80cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 22s then the acceleration of the stone is ?

Answer 1

We can start by using the formula for centripetal acceleration:

a = (v^2) / r

where a is the centripetal acceleration, v is the velocity of the stone, and r is the radius of the circle.

To find the velocity, we need to use the formula for angular velocity:

w = 2πn / t

where w is the angular velocity, n is the number of revolutions, and t is the time taken.

Plugging in the given values, we get:

w = (2π * 14) / 22 = 2.8π rad/s

To convert this to linear velocity (v), we can use the formula:

v = rw

where r is the length of the string. Plugging in the given value, we get:

v = 0.8 * 2.8π = 2.22 m/s

Now we can use the formula for centripetal acceleration:

a = (v^2) / r

Plugging in the values we calculated, we get:

a = (2.22^2) / 0.8 = 6.16 m/s^2

Therefore, the acceleration of the stone is 6.16 m/s^2.