Question

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?

Answer 1

The magnitude of the acceleration of the ball is 78.76 m/s², directed towards the center of the circle. The tension in the string is 13.77 N.

Step-by-step explanation:

To determine the acceleration of the ball, we need to first find its linear speed. The ball makes 2.0 revolutions every 1.0 s, so we can calculate its angular speed using the formula:

Angular speed = (2 * pi * revolutions) / time

angular_speed = (2 * 3.14 * 2.0) / 1.0 = 12.56 rad/s

Now, we can calculate the linear speed using the formula:

Linear speed = angular speed * radius

linear_speed = 12.56 * 0.5 = 6.28 m/s

The magnitude of the acceleration can be calculated using the formula:

Acceleration = (linear speed)^2 / radius

acceleration = (6.28)^2 / 0.5 = 78.76 m/s^2

Since the ball is moving in a horizontal circle, the acceleration is directed towards the center of the circle.

To calculate the tension in the string, we can use the centripetal force formula:

Tension = mass * acceleration

tension = 0.175 * 78.76 = 13.77 N

Answer 2

Explanation :

It is given that:

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

Radius of circle,
`r=(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* (4\pi)^2* 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*13.803\ m/s^2`

`F=2.415\ N`

This is the required answer.