. A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy? a. c. Zero d. 4T2R/T2 e. TR/T2

Respuesta :

Explanation:

Formula for centripetal acceleration of an object is as follows.

               a = [tex]\frac{v^{2}}{r}[/tex]

When an object is travelling in a circular path then it is difficult to measure its velocity.

Hence, for a circular object the formula for acceleration is as follows.

                a = [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]

     a = [tex]\frac{V^{2}}{r}[/tex],     and       V = [tex]\frac{d}{T} = \frac{2 \pi r}{T}[/tex]

     a = [tex]\frac{(\frac{[2\pi r]}{T})^{2}}{r}[/tex]

        = [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex]

Thus, we can conclude that the magnitude of the acceleration of the toy is [tex]\frac{4 \pi^{2} r}{T^{2}}[/tex].

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