Question

A small bead slides around a horizontal circle at height y inside the cone shown in FIGURE CP8.69. Find an expression for the bead’s speed in terms of a, h, y, and g.

Short answer

The expression for velocity is $v=\frac{a}{h}\sqrt{gy}$

Step-by-step solution

Given Information

A small bead slides around a horizontal circle at height y inside the cone .

Radius of base of cone = a

Height of cone = h

Explanation

Lets draw Free Body diagram as below

From the diagram equation vertical and horizontal components of force

$$\begin{gathered} R\sin \left(\theta \right)=mg..........................\left(1\right) \\ and, \\ Rcos\left(\theta \right)=\frac{m{v}^2}{r}................................\left(2\right) \end{gathered}$$

Divide equation(1) by (2) we get

$\frac{{\displaystyle \frac{m{v}^2}{r}}}{mg}=\left(\frac{R\cos \left(\theta \right)}{R\sin \left(\theta \right)}\right)\frac{v^2}{rg}=tan\left(\theta \right)..........................\left(3\right)$

From the diagram we can get angle

$$\begin{gathered} \frac{r}{y}=\frac{a}{h} \\ r=\frac{ay}{h} \\ \tan \left(\theta \right)=\frac{a}{h}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(4\right) \end{gathered}$$

From equation (3) and (4)

$$\begin{gathered} v^2=gr\tan \left(\theta \right) \\ {v}^2=g\left(\frac{ay}{h}\right)\left(\frac{a}{h}\right) \\ {v}^2=\frac{a^{2}gy}{h^2} \\ v=\frac{a}{h}\sqrt{gy} \end{gathered}$$