91影视

The passenger of mass$m$ is riding around a horizontal circle of radius$R$at speed$v$.

The problem deals with Newton鈥檚 laws of motion which describe the relations between the forces F, acting on a body and the motion of the body. Also, it deals with centripetal acceleration. This is the acceleration, a of a body traversing a circular path of radius R.

Formula:

According to Newton鈥檚 laws of motion,

$\mathrm{贵蝉颈苍胃}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

$\mathrm{贵肠辞蝉胃}=\mathrm{mg}$

Centripetal acceleration is given by,

$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$

From the free body diagram, note that $\mathrm{F}$is the force of aerodynamic lift and$\mathrm{a}$is towards the right.

Centripetal acceleration,

$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ (i)

Applying Newton鈥檚 law,

$\mathrm{贵蝉颈苍胃}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$ (ii)

$\mathrm{贵肠辞蝉胃}=\mathrm{mg}$ (iii)

Eliminating mass from equations (ii) and (iii) as:

$\mathrm{迟补苍胃}=\frac{{\mathrm{v}}^2}{\mathrm{gR}}$

We have,

$\begin{array}{c}\mathrm{v}=480\text{鈥塳尘/丑}\\ =133\text{鈥尘/s}\end{array}$

and $\mathrm{胃}={49}^{\mathrm{o}}$

Then,

$\begin{array}{c}\mathrm{R}=\frac{{\mathrm{v}}^2}{\mathrm{g迟补苍胃}}\\ =\frac{{133}^2}{9.8脳\tan {40}^{\text{o}}}\\ =2151\text{鈥尘}\end{array}$

Thus, radius of the circle is ,$\mathrm{R}=2.2脳{10}^3\text{鈥尘}$.