Question

A body initially at rest and sliding along a frictionless track from a height $h$ just completes a vertical circle of diameter $AB=D$. What is the height h equal to?

Answer 1

Step1: Given data

A body initially at rest and sliding along a frictionless track from a height $h$

A vertical circle of diameter $AB=D$.

Step2: Formula used

Potential energy, $P.E.=mgh$

Kinetic energy, $K.E.=\frac{1}{2}m{v}^2$

Step3: Calculating the height

The energy at the top= The energy at the bottom
At the top, there is only potential energy and at the bottom, the energy is completely kinetic, so,

$$\begin{gathered} mgh=\frac{1}{2}m{v}^2 \\ h=\frac{v^2}{2g} \end{gathered}$$

For completing the vertical circle, the velocity at the lowest point must satisfy the condition,

$v\ge \sqrt{5gr}$ where $r$ is the radius and equal to half of the diameter.

Now, the value of h becomes,

$$\begin{gathered} h=\frac{5gr}{2g} \\ h=\frac{5}{4}D \end{gathered}$$

Hence, the height h is equal to $\frac{5}{4}D$.