Question

. 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

Answer 1

Step-by-step explanation:

Formula for centripetal acceleration of an object is as follows.

a =
`(v^(2))/(r)`

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 =
`(4 \pi^(2) r)/(T^(2))`

a =
`(V^(2))/(r)`, and V =
`(d)/(T) = (2 \pi r)/(T)`

a =
`((([2\pi r])/(T))^(2))/(r)`

=
`(4 \pi^(2) r)/(T^(2))`

Thus, we can conclude that the magnitude of the acceleration of the toy is
`(4 \pi^(2) r)/(T^(2))`.