A small bead can slide without friction on a circular hoop that is in a vertical plane and has a radius of 0.100 m. The hoop rotates at a constant rate of 4.00 rev/s about a vertical diameter (Fig. P5.107). (a) Find the angle$\mathit{\beta }$ at which the bead is in vertical equilibrium. (It has a radial acceleration toward the axis.) (b) Is it possible for the bead to “ride” at the same elevation as the center of the hoop? (c) What will happen if the hoop rotates at 1.00 rev/s?

According to Newton’s second law, the linear force is given by,

$F=ma$

Here, F is linear force, m is the mass of object, and a is acceleration of object.

The centripetal acceleration is given by,

$a_c=\frac{v^2}{r}$

Here, is velocity, and is radius of curvature.

(a)

Draw the free-body diagram of the bead.

The Newton’s equation of motion of the system at equilibrium is given by,

$\sum F_y=m{a}_y......\left(1\right)$

Here,$F_y$is the vertical force on the bead, m is the mass of the bead, and$a_y$is vertical acceleration.

Rewrite the vertical component of the force as $a_y=0$.

$n\cos \beta -mg=0......\left(2\right)$

Here, n is normal force on the bead, and$\beta$is the angle of bead from the vertical axis of the hoop.

Write the vertical component of the force as $a_x=0$.

$$\begin{gathered} \sum F_x=m{a}_x \\ n\sin \beta -m{a}_{\text{rad}}=0......\left(3\right) \end{gathered}$$

Here,$a_{\text{rad}}$is the radial acceleration of the bead.

The radial acceleration is given by,

$a_{\text{rad}}=\frac{v^2}{R}......\left(4\right)$

Here, v is linear velocity of the bead, and R is the horizontal distance from the vertical rotating axis of the hoop.

The expression of linear speed is given by,

$v=\frac{2\pi R}{T}......\left(5\right)$

Here, T is time for one revolution.

From equations (2), (3), (4), and (5),

$\beta ={\cos }^{-1}\left(\frac{T^{2}g}{4{\pi }^{2}r}\right)......\left(6\right)$

here, r is the radius of hoop.

Substitute$\frac{1}{4}\text{s}$ for T,$9.8m/s^2$ for g, 0.100m for r in equation (6).

$\begin{array}{rcl}\beta & =& {\cos }^{-1}\left(\frac{{\left(\frac{1}{4}\text{s}\right)}^{2}9.8m/s^2}{4{\pi }^2\left(0.100\text{m}\right)}\right)\\ & =& 81.1°\end{array}$

Therefore, the required angle$\beta$ is $81.1°$.

(b)

It is impossible for the bead to ride at the same elevation because it means that $\cos \beta =0$, and if $\cos \beta =0$then r or g will be equal to zero which is impossible at all.

Therefore, no, it is not possible for the bead to “ride” at the same elevation as the center of the hoop.

(c)

From the law of centripetal force,

$\begin{array}{rcl}{F}_c& =& \frac{m{v}^2}{r}\\ & =& m{\omega }^{2}r\\ & =& m{\left(2\pi N\right)}^{2}r\end{array}$

From the above relation, it is clear that $F_c\propto {\left(N\right)}^2$, so if N becomes smaller then the value of centripetal force must become smaller, and the vertical component $F_{cy}$of the force will be larger and the ball will be closer to the bottom of the hoop.

Therefore, the ball will be closer to the bottom of the hoop if the hoop rotates at $1.00rev/s$.