A high-speed railway car goes around a flat, horizontal circle of radius $\mathbf{470}\mathbf{m}$at a constant speed. The magnitudes of the horizontal and vertical components of the force of the car on a$\mathbf{51}\mathbf{.}\mathbf{0}\mathbf{kg}$passenger are$\mathbf{210}\mathbf{N}$and$\mathbf{500}\mathbf{N}$, respectively.

(a)What is the magnitude of the net force (of allthe forces) on the passenger?

(b) What is the speed of the car?

• The radius of the circle,$\mathrm{R}=470\text{\hspace{0.17em}m}$.
• The horizontal component of force,${\mathrm{F}}_{\mathrm{h}}=210\text{\hspace{0.17em}N}$.
• The vertical component of force,${\mathrm{F}}_{\mathrm{v}}=500\text{\hspace{0.17em}N}$.
• Mass of the passenger, $\mathrm{m}=51\text{kg}$.

The railway car is moving around the horizontal circle at a constant speed. Thus, the horizontal component of the force must supply the centripetal force, which is necessary for maintaining the circular motion.

Formula:

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

The upward force exerted by the car on the passenger is equal to the downward force of gravity$(\mathrm{W}=500\text{N})$on the passenger. So the net force does not have a vertical contribution; it only has the contribution from the horizontal force (which is necessary for maintaining the circular motion). Thus,

$\begin{array}{c}\left|{\overrightarrow{\mathrm{F}}}_{\mathrm{net}}\right|=\mathrm{F}\\ =210\text{N}\end{array}$

Thus, the magnitude of the net force on the passenger is 210 N.

Centripetal force is given by,

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

$\mathrm{v}=\sqrt{\frac{\mathrm{FR}}{\mathrm{m}}}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{\left(210\text{\hspace{0.17em}N}\right)\left(470\text{\hspace{0.17em}m}\right)}{51.0\text{\hspace{0.17em}kg}}}\\ \mathrm{v}=44.0\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed of the car is 44 m/s.