Question

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?

Answer 1

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

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

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$