An athlete whirls a 7 kg hammer tied to the end of a 1.3 m long chain in a horizontal circle. The hammer makes one revolution in 1 s.What is the centripetal force (and direction!)acting on the hammer

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Answer:

The centripetal force acting on the hammer is 57.14 N.

Explanation:

Given that,

Mass of the hammer, m = 7 kg

Length of the chain, l = 1.3 m

The hammer makes one revolution in 1 s. The length of the chain will be the radius of the circular path. So, the formula of centripetal force is given by :

[tex]F=mr\omega^2[/tex]

Here, [tex]\omega=6.28\ rad/s[/tex]

[tex]F=7\times 1.3\times 6.28\\\\F=57.14\ N[/tex]

So, the centripetal force acting on the hammer is 57.14 N. Hence, this is the required solution.

The centripetal force acting on the hammer is 358.9 N and it is directed inwards.

The given parameters:

  • mass of the hammer, m = 7 kg
  • radius of the circle, r = 1.3 m
  • time to make one revolution, t = 1 s

The angular speed of the hammer is calculated as follows;

[tex]\omega = 1 \ \frac{rev}{1 \ s} \times \frac{2\pi \ rad}{rev} \\\\\omega = 6.28 \ rad/s[/tex]

The centripetal force acting on the hammer is calculated as follows:

[tex]F = ma_c\\\\F = m \omega ^2r\\\\F = 7 \times (6.28)^2 \times 1.3\\\\F = 358.9 \ N[/tex]

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