Question

A child sitting 1.20 m from the center of a merry go round moves with a speed of 1.10m/s. Calculate (a) the centripetal acceleration of the child and (b) the net horizontal force exerted on the child (mass=22.5kg)

Answer 1

Centripetal acceleration = 1 and force is 22.5

Explanation: centripetal acceleration= v^2/R=1 and force = m ×centripetal acceleration = 22.5

Answer 2

The centripetal acceleration of the child and the net horizontal force exerted on the child are 1m/s² and 22.5N respectively.

Given:-

Radius of Merry go round (r) = 1.20m

Speed of Merry go round (v) = 1.10m/s

Mass of Child (m) = 22.5kg

To Find:-

The centripetal acceleration of the child and the net horizontal force exerted on the child.

Solution:-

We can simply find out the centripetal acceleration of the child and the net horizontal force exerted on the child by using the following procedure.

As,

Radius of Merry go round (r) = 1.20m

Speed of Merry go round (v) = 1.10m/s

Mass of Child (m) = 22.5kg

According to the formula,

a). Centripetal acceleration,

$ac = \frac{ {v}^{2} }{r}$

$ac = \frac{ {1.1}^{2} }{1.2}$

$ac = \frac{1.21}{1.20}$

ac = 1m/s²

b). Net horizontal force on the Child.

$force = mass \times acceleration$

$f = m \times ac$

$f = 22.5 \times 1$

f = 22.5N

Hence, The centripetal acceleration of the child and the net horizontal force exerted on the child are 1m/s² and 22.5N respectively.

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