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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. What are the magnitude and direction of the acceleration of the ball?


What's the tension in the string?

Verified Answer :

olemakpadu

For circular motion.

Centripetal acceleration = mv²/r = mω²r

Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m

m = mass = 175g = 0.175kg.

Angular speed, ω = Angle covered / time

                         = 2 revolutions / 1 second

                         = 2 2π  radians / 1 second

                         = 4π  radians / second

Centripetal Acceleration = mω²r = 0.175(4π)² 0.5    Use a calculator

                                                         ≈13.817  m/s²

The magnitude of acceleration ≈13.817  m/s² and it is directed towards the center of rotation.

Tension in the string = ma

                                   = 0.175*13.817

                                   = 2.418 N

Answer :

Explanation :

It is given that:

mass of the ball, \(m=175\ g=0.175\ Kg\)

Radius of circle, \(r=\dfrac{diameter}{2}=0.5\ m\)

The ball makes 2.0 revolutions every 1.0 s. So, angular speed is \(\omega=4\pi\ radian/sec\)

Since, it is moving in circular path so centripetal acceleration will act here.

So, centripetal acceleration \(\alpha\) =\(m\omega^2r\)

\(\alpha=0.175\ Kg\times (4\pi)^2\times 0.5\)

So, \(\alpha=13.803\ m/s^2\)

Hence, the acceleration is \(13.803\ m/s^2\) and it is directed towards the center of rotation.

Tension is a force which is given by :

                                 \(F=ma\)

                        \(F=0.175\ Kg\times13.803\ m/s^2\)

                                   \(F=2.415\ N\)

This is the required answer.                        

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