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An athlete whirls an 8.2 kg hammer tied to the end of a 1.4 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 1.85 rev/s. What is the centripetal acceleration of the hammer? Assume his arm length is included in the length given for the chain. Answer in units of m/s 2 .

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Answer:

Explanation:

mass of hammer, m = 8.2 kg

Length of chain, l = 1.4 m

angular frequency, f = 1.85 rev/s

Let the centripetal acceleration is a.

a = r x ω²

a = l x 4π² x f²

a = 1.4 x 4 x 3.14 x 3.14 x 1.85 x 1.85

a = 188.97 m/s²

boffeemadrid

The centripetal acceleration of the given mass is required.

The centripetal acceleration is [tex]189.16\ \text{m/s}^2[/tex]

r = Radius = 1.4 m

[tex]\omega[/tex] = Angular velocity = 1.85 rev/s

Centripetal acceleration is given by

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{(\omega r)^2}{r}\\\Rightarrow a_c=\omega^2 r\\\Rightarrow a_c=(1.85\times 2\pi)^2\times 1.4\\\Rightarrow a_c=189.16\ \text{m/s}^2[/tex]

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