91Ó°ÊÓ

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,

$$\begin{gathered} \underset{}{∑}F=-m{a}_c \\ R-mg=-m{a}_c \end{gathered}$$

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$9.8 \mathrm{m}/{\mathrm{s}}^2$ 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

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

Substitute$9.8 \mathrm{m}/{\mathrm{s}}^2$ 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).

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

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