​ A small body tied to a string is revolved in a vertical circle of radius r such that its speed at the top of the circle is √(2gr). Find ​

Question

A small body tied to a string is revolved in a vertical circle of radius r such that its speed at the top of the circle is \(\sqrt{2gr}\). Find 

(i) the angular position of the string when the tension in the string is numerically equal to 5 times the weight of the body. 

(ii) the KE of the body at this position 

(iii) the minimum and maximum KEs of the body.

[Take m = 0.1 kg, r = 1.2 m, g = 10 m/s^s]

Answer 1

Data : v_top = \(\sqrt{2gr}\), T = 5 mg, m = 0.1 kg, r = 1.2 m, g = 10 m/s²

Let the angular position of the string, θ = 0° when the body is at the bottom of the circle.

We assume total energy to be conserved and take the reference level for zero potential energy to be the bottom of the circle.

Total energy at the top, E

(i) At position P where T = 5 mg.

At P, the vertical displacement of the body from the bottom is r(1 – cos θ). Its total energy there is also E.

Substituting for v² from Eq.(2) and simplifying

(ii) At p, v² = gr(5 - cos \(\theta\))

(iii) The body has minimum KE at the top and maximum KE at the bottom.