Question

A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m above the ground. The string breaks and the ball lands 1.60 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

Answer:

radial acceleration is 41.8 m / s²

Explanation:

The acceleration for circular motion is

     a = v² / r

They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations

     x = vox t

     y = v₀ₓ t - ½ g t2

Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)

     x = v₀ₓ t

     t = x / v₀ₓ

     y = - ½ g t²

Let's replace and calculate the initial velocity on the X axis

    y = - ½ g (x / vox)²

    v₀ₓ = √ (g x² / 2 y)

    v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]

    v₀ₓ = 3.54 m / s

This is the horizontal velocity, but since it circle is in  horizontal position it is also the velocity of the body at the point of rupture.

Now we can calculate the radial acceleration

        a = v² / r

       a = 3.54² / 0.300

       a = 41.8 m / s²