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

Answer 1

We can start by using the formula for centripetal acceleration:

a = (v^2) / r

where v is the speed of the stone and r is the length of the string. We can also use the formula for the period of circular motion:

T = 2πr / v

where T is the time for one revolution. We know that the stone makes 14 revolutions in 25 s, so the period is:

T = 25 s / 14 = 1.79 s/rev

Using this information, we can solve for the speed of the stone:

v = 2πr / T = 2π(80 cm) / 1.79 s = 282.7 cm/s

Now we can use the formula for centripetal acceleration:

a = (v^2) / r = (282.7 cm/s)^2 / 80 cm = 996.8 cm/s^2

The magnitude of the acceleration is 996.8 cm/s^2. To find the direction, we can use the fact that the acceleration is always directed towards the center of the circle. In this case, since the motion is horizontal, the acceleration is also horizontal and towards the center.

Therefore, the magnitude of acceleration is 996.8 cm/s^2 and the direction is horizontal and towards the center of the circle.