**Question:**A tether-ball attached to the end of a 4.00 m rope is swung in a horizontal circular path around a child’s head. The mass of the tether-ball is 1.00 kg. If the tether-ball rotates at a rate of 45.0 revolutions per minute, what is the centripetal acceleration of the ball?**Options:**- 11.3 $ \text{m/s}^2 $- 8100 $ \text{m/s}^2 $- 88.8 $ \text{m/s}^2 $- 29.6 $ \text{m/s}^2 $- 47.0 $ \text{m/s}^2 $**Instructions:**Choose the correct option that represents the centripetal acceleration of the ball.

Answered: A tether-ball attached to the end of a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Given Data:

The mass of a tether-ball is, $m = 1.00 kg$

The angular speed of a tether-ball is, $ω = 45.0 rpm$

The radius of a circular path followed by a rope is, $r = 4.00 m$

The expression for the centripetal acceleration is given as,

$\begin{array}{rcl} a_{c} & = & \frac{v^{2}}{r} (∴ v = r ω) \\ a_{c} & = & \frac{{(r ω)}^{2}}{r} \\ a_{c} & = & ω^{2} r . . . . . . . . . . . . . . . . . (1) \end{array}$

Here, $ω$ is the angular speed, and r is the radius of a circular path.

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