Question

A bead slides without friction around a loop-the-loop (Fig. P8.5). The bead is released from rest at a height $\mathit{h}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathit{R}$. (a) What is its speed at point A? (b) How large is the normal force on the bead at point A if its mass is $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\text{g}}$?

Short answer

(a) The speed of bead at point A is ${\left(3gR\right)}^{1/2}$.

(b) The normal force on bead at point A is $0.098\text{N}$.

Step-by-step solution

A concept and identification of given data

The principle of energy conservation is used find the speed of bead at point A. The normal force on the bead is equal to the weight of bead.

Consider the given data as below.

The mass of bead, $m=5\text{g}$

The initial position of bead, $h=3.50R$

As known that, the acceleration due to gravity is

$g=9.8m/s^2$

(a) Determination of speed of bead at point A

Apply the work energy theorem to find the speed of bead at bottom of loop.

$\frac{1}{2}m\left(v^2-u^2\right)=mgh$

Here, $g$ is the gravitational acceleration and $u$ is the initial speed of bead and its value is zero.

Substitute all the values in the above equation.

$$\begin{gathered} \frac{1}{2}m\left(v^2-{\left(0\right)}^2\right)=mg\left(3.50R\right) \\ v=\sqrt{7gR} \end{gathered}$$

Apply the energy conservation between bottom of loop and point A.

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_A^2+mg{h}_A$

Here, is the distance of point A from bottom of loop.

Substitute all the values in the above equation.

$$\begin{gathered} \frac{1}{2}m{\left(\sqrt{7gR}\right)}^2=\frac{1}{2}m{v}_A^2+mg\left(2R\right) \\ \frac{1}{2}{v}_A^2=3.50gR-2gR \\ {v}_A={\left(3gR\right)}^{1/2} \end{gathered}$$

Therefore, the speed of bead at point A is ${\left(3gR\right)}^{1/2}$.

(b) Determination of normal force on bead at point A

The normal force on bead at point A is given as

$N=2mg$

Substitute all the values in the above equation.

$$\begin{gathered} N=2\left(5g\right)\left(\frac{1kg}{1000g}\right)\left(9.8m/s^2\right) \\ =0.098N \end{gathered}$$

Hence, the normal force on bead at point A is 0.098 N.