A ball on the end of a string is whirled around in a horizontal circle of radius 0.386 m. The plane of the circle is 1.10 m above the ground. The string breaks and the ball lands 2.00 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion.

Question

A ball on the end of a string is whirled around in a horizontal circle of radius 0.386 m. The plane of the circle is 1.10 m above the ground. The string breaks and the ball lands 2.00 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion.

Answer 1

To find the radial acceleration of the ball during its circular motion, we can use the centripetal acceleration formula:

a = (v^2) / r

where a is the radial acceleration, v is the velocity of the ball, and r is the radius of the circle.

First, we need to find the velocity of the ball just before the string breaks. We can relate the horizontal distance traveled by the ball (2.00 m) to its vertical distance above the ground (1.10 m) using the Pythagorean theorem.

Let's consider the triangle formed by the vertical displacement (h), the horizontal distance (d), and the radius of the circle (r) as the hypotenuse. From the given information, we have:

d^2 = r^2 + h^2

Substituting the values, we have:

(2.00 m)^2 = (0.386 m)^2 + (1.10 m)^2

Simplifying,

4.00 m^2 = 0.149 m^2 + 1.21 m^2

Combining like terms,

4.00 m^2 = 1.359 m^2

Taking the square root of both sides,

2.00 m = √1.359 m

Therefore, the distance traveled in the vertical direction before the string breaks is 2.00 m.

Now, we can find the velocity of the ball just before the string breaks. We know that the time it takes for the ball to travel the vertical distance (1.10 m) is the same as the time it takes to travel the horizontal distance (2.00 m). Using the formula for time, t = d / v, we can set up the following equation:

1.10 m / v = 2.00 m / v

Simplifying,

1.10 m = 2.00 m

Since the distances are equal, the time is also equal. Therefore, we can conclude that the velocity is the same just before the string breaks.

Now, we can use the given radius of the circle (0.386 m) and the velocity of the ball to calculate the radial acceleration. Let's substitute the values into the formula:

a = (v^2) / r

Using the velocity of the ball, we have:

a = (v^2) / r = (2.00 m/s)^2 / 0.386 m

Calculating,

a ≈ 10.36 m/s^2

Therefore, the radial acceleration of the ball during its circular motion is approximately 10.36 m/s^2.