Solved Problem on Dynamics
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A high-speed train travels a 2500 m radius curve at a speed of 270 km/h. Determine:
a) The centrifugal force felt by a passenger of a mass of 70 kg in a wagon;
b) What should be the speed of a car traveling a 10 m radius curve so that the passenger of the train feels the same centrifugal force being in the car? Answer in km/h.


Problem data:
  • Radius of the curve of the train:    rT = 2500 m;
  • Speed of the train:    vT = 270 km/h;
  • Passenger mass:    m = 70 kg;
  • Radius of the curve of the car:    rC = 10 m.
Problem diagram:

When the train turns, the centripetal acceleration acts on the train and the bodies inside (passengers and loads). This acceleration is responsible for making the bodies go through the curve. In a reference frame on the train, the bodies feel the centrifugal force that balances the centripetal force (Figure 1).

Figure 1

Note: the centripetal force only changes the direction of movement and not its tangential speed. The train continues at the same speed of 270 km/h.

Solution

First, we convert the given train speed in kilometers per hour (km/h) to meters per second (m/s) used in the International System of Units (SI)
\[ \begin{gather} v_{\small T}=270\;\frac{\mathrm{\cancel{km}}}{\mathrm{\cancel{h}}}\times\frac{1000\;\mathrm m}{1\;\mathrm{\cancel{km}}}\times\frac{1\;\mathrm{\cancel{h}}}{3600\;\mathrm s}=\frac{270}{3.6}\;\frac{\mathrm m}{\mathrm s}=75\;\mathrm{m/s} \end{gather} \]
a) Using Newton's Second Law for a motion in a curve
\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec F}_{cp}=m {\vec a}_{cp}} \tag{I} \end{gather} \]
the centripetal acceleration is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {a_{cp}=\frac{v^2}{r}} \tag{II} \end{gather} \]
substituting the equation (II) into equation (I)
\[ \begin{gather} F_{cp}=m \frac{v^2}{r} \tag{III} \end{gather} \]
Applying equation (III) to the train passenger
\[ \begin{gather} F_{cp}=m\frac{v_{\small T}^{2}}{r_{\small T}}\\[5pt] F_{cp}=\left(70\;\mathrm{kg}\right)\frac{\left(75\;\mathrm{\frac{m}{s}}\right)^2}{2500\;\mathrm m}\\[5pt] F_{cp}=\left(70\;\mathrm{kg}\right)\frac{\left(5625\;\mathrm{\frac{m^{\cancel 2}}{s^2}}\right)}{2500\;\mathrm{\cancel{m}}}\\[5pt] F_{cp}=157.5\ \mathrm N \end{gather} \]
as the centripetal force and centrifugal force must be equal in magnitude
\[ \begin{gather} F_{cp}=F_{cg} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{cg}=157.5\;\mathrm N} \end{gather} \]

b) If the train passenger is in a car, the centripetal force (and the centrifugal force) acting on it shall be by applying (III)
\[ \begin{gather} F_{cp}=F_{cg}=m\frac{v_{\small C}^2}{r_{\small C}}\\[5pt] v_{\small C}=\sqrt{{\frac{ F_{cg}r_{\small C} }{m}}\;}\\[5pt] v_{\small C}=\sqrt{\frac{\left(157.5\;\mathrm N\right)\left(10\;\mathrm m\right)}{70\;\mathrm{kg}}\;}\\[5pt] v_{\small C}=4.7\; \mathrm{m/s} \end{gather} \]
Figure 2

Converting the answer to km/h
\[ \begin{gather} v_{\small C}=4.7\;\mathrm{\frac{\cancel{m}}{\cancel{s}}}\times\frac{1\;\mathrm{km}}{1000\;\mathrm{\cancel{m}}}\times\frac{3600\;\mathrm{\cancel{s}}}{1\;\mathrm h}=16.9\; \mathrm{km/h} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{\small C}\;\approx \;17\; \mathrm{km/h}} \end{gather} \]
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