Question

A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attached to a frictionless pivot. The rod is raised until it is vertical, with the stone above the pivot. The rod is released and the stone moves in a vertical circle with no air resistance. What is the tension in the rod as the stone moves through the bottom of the circle? (a) 20 N (b) 40 N (c) 60 N (d) 80 N (e) 100 N

Answer 1

Answer:

Option B is the correct answer.

Explanation:

Refer the figure we have centripetal force at bottom of circle

       $F_c=F_t-F_w\\\\\frac{mv^2}{r}=F_t-mg\\\\F_t=m\left ( \frac{v^2}{r}+g\right )$

We have mass, m = 2 kg

Radius, r = 1.2 m

For circular motion to occur we have tension at top = 0

That is

        $\frac{mv^2}{r}=mg\\\\v=\sqrt{rg}$

Now let us find tension at bottom point

       $F_t=2\times \left ( \frac{rg}{r}+g\right )=4g=40N$

Option B is the correct answer.