A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius 13.0 m. She has mass 70.0 kg, and her motorcycle has mass 40.0 kg. (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

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

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here,F is linear force,m is the mass of object, anda 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 given situation.

According to the Newton’s second Law, the net force at the top is given by,

$$\begin{gathered} \sum _{}F=-m{a}_c \\ R-mg=-m{a}_c \end{gathered}$$

To loose the contact means R=0 . Then

$$\begin{gathered} 0-mg=-m\left(\frac{v^2}{r}\right) \\ {v}^2=gr \\ {v}_{min}=\sqrt{gr}.........\left(1\right) \end{gathered}$$

Substitute$9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$ for g,and 13 m forr in equation (1).

$$\begin{gathered} v_{min}=\sqrt{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(13\mathrm{m}\right)} \\ =\sqrt{127.4} \\ =11.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the minimum velocity is 11.3 m/s.

(b)

The net force acting on the motor cycle is given by,

$$\begin{gathered} R_m-m_{m}g-m_{p}g=m_m{a}_c \\ {R}_m=g\left(m_m+m_p\right)+\frac{m_m{v}^2}{r}............\left(2\right) \end{gathered}$$

Here,$R_m$ is the apparent force by the sphere on motor,$R_p$ is the reaction force acting on sphere due to professor weight.

The velocity is twice in this case, then

$$\begin{gathered} \mathrm{v}=2\times 11.3\mathrm{m}/\mathrm{s} \\ =22.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute$9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$ for g, 40 kg for$m_m$, 70 kg for$m_p$,$22.6\mathrm{m}/\mathrm{s}$ for v,and 13 m forr in equation (2).

$$\begin{gathered} R_m=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(40\mathrm{kg}+70\mathrm{kg}\right)+\frac{\left(40\mathrm{kg}\right){\left(22.6\mathrm{m}/\mathrm{s}\right)}^2}{13\mathrm{m}} \\ =1079.1+1571.57 \\ \approx 2651\mathrm{N} \end{gathered}$$

Therefore, the required magnitude of the normal force is 2651 N.