Question

A body of mass 2 kg lying on a smooth surface attached to a string 3 m long and then rotated in a horizontal circle making 60 rev/min.Calculate the centripetal force​

Answer 1

Answer:

Approximately 237 N

Explanation:

F = mrω²

= mr(2πn)²

= 2 kg × 3 m × (2πn)²

= 6 kg m × (2 × 3.14 × 60 rev / 60 s)²

= 6 kg m × 6.28² /s²

≈ 237 kg m/s²

≈ 237 N

Answer 2

The centripetal force acting on the body is 236.63 N.

Explanation:

It is given that,

Mass of the body, m = 2 kg

Length of the string, l = r = 3 m

Angular speed of the body, $\omega=60\ rev/min=6.28\ rad/s$

Let F is the centripetal force acting on the body. It is given by :

$F=mr\omega^2$

$F=2\ kg\times 3\ m\times (6.28\ rad/s)^2$

F = 236.63 N

So, the centripetal force acting on the body is 236.63 N. Hence, this is the required solution.

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