91影视

According to Newton鈥檚 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,

${\mathit{a}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here, v is velocity, and r is radius of curvature.

(a)

Draw the free-body diagram of the bead.

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

$\underset{}{鈭�}{F}_y=m{a}_y$ ........(1)

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 $a_y=0$.

$n\cos 尾-mg=0........\left(2\right)$

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

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

$$\begin{gathered} \underset{}{鈭�}{F}_x=m{a}_x \\ n\sin 尾-m{a}_{rad}=0.......\left(3\right) \end{gathered}$$

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

The radial acceleration is given by,

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

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,

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

Here, T is time for one revolution.

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

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

here , r is the radius of hoop.

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

$$\begin{gathered} \mathrm{尾}={\cos }^{-1}\left(\frac{{\left({\displaystyle \frac{1}{4}}\mathrm{s}\right)}^2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{4{\mathrm{蟺}}^2\left(0.100\mathrm{m}\right)}\right) \\ =81.1掳 \end{gathered}$$

Therefore, the required angle$\mathrm{尾}$ is$81.1掳$ .

(b)

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

Therefore, no, it is not possible for the bead to 鈥渞ide鈥� at the same elevation as the center of the hoop.

(c)

From the law of centripetal force,

$$\begin{gathered} F_c=\frac{m{v}^2}{r} \\ =m{蝇}^{2}r \\ =m{\left(2蟺N\right)}^{2}r \end{gathered}$$

From the above relation, it is clear that $F_c鈭�{\left(N\right)}^2$, so if 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 more closer to the bottom of the hoop if the hoop rotates at 1.00 rev/s .