A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with aconstant speed. If the stone makes 8 revolutions in 20 s, what is the magnitude anddirection of acceleration of the stone?
Question
A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with aconstant speed. If the stone makes 8 revolutions in 20 s, what is the magnitude anddirection of acceleration of the stone?
Solution
To solve this problem, we need to find two things: the speed of the stone and the centripetal acceleration.
Step 1: Find the speed of the stone First, we need to find the distance the stone travels in one revolution. This is the circumference of the circle, which is 2πr. The radius r is the length of the string, which is 100 cm or 1 meter. So, the distance for one revolution is 2π(1) = 2π meters.
The stone makes 8 revolutions in 20 seconds, so the total distance it travels is 8(2π) = 16π meters.
The speed of the stone is the total distance divided by the time, so the speed is 16π/20 = 0.8π meters per second.
Step 2: Find the centripetal acceleration The formula for centripetal acceleration is a = v²/r, where v is the speed and r is the radius. Plugging in the values we found, the acceleration is a = (0.8π)²/1 = 0.64π² meters per second squared.
Step 3: Find the direction of the acceleration The direction of the acceleration is always towards the center of the circle. So, if the stone is being whirled around, the acceleration is towards the hand that is swinging the stone.
So, the magnitude of the acceleration is 0.64π² meters per second squared, and the direction is towards the center of the circle.
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