You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m in diameter. What is the minimum speed the rock must have at the top of the circle if it to always stay-in contact with the bottom of the bucket?

Respuesta :

Answer:v=3.28 m/s

Explanation:

Given

mass of rock [tex]m=500 gm[/tex]

diameter of circle [tex]d=2.2 m[/tex]

radius [tex]r=\frac{2.2}{2}=1.1 m[/tex]

At highest Point

[tex]mg+N=\frac{mv^2}{r}[/tex]

At highest Point N=0 because mass is just balanced by centripetal Force

thus [tex]mg=\frac{mv^2}{r}[/tex]

[tex]v=\sqrt{gr}[/tex]

[tex]v=\sqrt{9.8\times 1.1}[/tex]

[tex]v=\sqrt{10.78}[/tex]

[tex]v=3.28 m/s[/tex]

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