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 rate of 45.0 revolutions per minute, what is centripetal acceleration of the ball?

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 rate of 45.0 revolutions per minute, what is centripetal acceleration of the ball?

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter6: Motion In Two Dimensions

Section6.2: Circular Motion

Problem 13PP

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

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 rate of 45.0 revolutions per minute, what is centripetal acceleration of the
ball?
O 11.3 m/s?
8100 m/s?
88.8 m/s?
29.6 m/s?
O 47.0 m/s?

Expert Solution

Step 1

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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