A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame

Question

3 months ago

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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 diame

ter 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?

Physics

2 answers:

Yuliya22 [3.3K] · Answer 1 · 2026-03-23T12:56:01+03:00

Answer:

Explanation:

Given are:

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

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

The ball completes 2.0 revolutions in every 1.0 s, thus yielding an angular speed of $\omega=4\pi\ radian/sec$

As it travels in a circular trajectory, centripetal acceleration will be exerted here.

Thus, centripetal acceleration $\alpha$ = $m\omega^2r$

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

Consequently, $\alpha=13.803\ m/s^2$

Therefore, the acceleration is $13.803\ m/s^2$ and it points towards the center of rotation.

Tension is a force defined by:

$F=ma$

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

$F=2.415\ N$

This represents the needed answer.

kicyunya [3.2K] · Answer 2 · 2026-03-23T12:55:39+03:00

For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

   = 2 revolutions per 1 second

   = 2 * 2π radians for each second

   = 4π radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

   ≈13.817 m/s²

. The acceleration's magnitude is approximately 13.817 m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

   = 0.175*13.817

   = 2.418 N