A ball on the end of a string is whirled around in a horizontal circle of radius 0.273 m. The plane of the circle is 1.31 m above the ground. The string breaks and the ball lands 2.87 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9.8 m/s 2 . Find the centripetal acceleration of the ball during its circular motion. Answer in units of m/s 2

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-11-18T03:34:28+01:00

Answer:

[tex]a_c=112.85 m/s^2[/tex]

Explanation:

The centripetal acceleration is:

[tex]a_c=\frac{v^2}{R}[/tex]

[tex]y_f=\frac{1}{2}*a*t^2[/tex]

The acceleration on this axis is g and the y' is the distance above the ground so solve to t'

[tex]t=\sqrt{\frac{2*y_f}{g}}=\sqrt{\frac{2*1.31m}{9.8m/s^2}}[/tex]

[tex]t=0.517s[/tex]

Now the distance and the time take the ball lands, can find the speed using

[tex]v=\frac{x}{t}=\frac{2.87m}{0.517s}[/tex]

[tex]v=5.55m/s[/tex]

So finally the centripetal acceleration is

[tex]a_c=\frac{v^2}{r}=\frac{(5.5m/s)^2}{0.273m}[/tex]

[tex]a_c=112.85 m/s^2[/tex]