Roller coaster

We consider a roller coaster with a lift hill of height h = 30 m and a circular looping with radius R = 10 m. We assume, that the cars are only accelerated by gravity. If the cars start with zero velocity at point A, how large are the absolute values of the g-force in point B and C?

Additional question: A vertical g-force up to 5g is for the human body easily bearable. Can you modify the roller coaster, so that the maximal g-force is below this value (without removing the looping)?

2g and 6g 6g and 6g 7g and 1g 6g and 2g

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1 solution

The total energy

E = U + T = m g y + 1 2 m v 2 E = U + T = m g y + \frac{1}{2} m v^2 (potential and kinetic energy)

of a car with height y y , velocity v v and mass m m is conserved along the track. The energy in the points A, B and C reads:

E A = m g h E B = 1 2 m v B 2 E A = E B v B = 2 g h E C = 1 2 m v C 2 + 2 m g R E A = E C v C = 2 g ( h 2 R ) \begin{aligned} E_A &= m g h & & \\ E_B &= \frac{1}{2} m v_B^2 & E_A = E_B \quad \Rightarrow \quad & v_B = \sqrt{2 g h} \\ E_C &= \frac{1}{2} m v_C^2 + 2 m g R & E_A = E_C \quad \Rightarrow \quad & v_C = \sqrt{2 g (h - 2 R)} \end{aligned}

With the help of the velocities v B v_B and v C v_C we can caculate the centrifugal force F cf = ± m v 2 / R F_\text{cf} = \pm m v^2/R inside the looping. With the gravitional force F g = m g F_\text{g} = -mg we get the total force, that acts on the passengers:

F B = m v B 2 R m g = ( 2 h R + 1 ) m g = 7 m g F C = + m v C 2 R m g = + ( 2 h R 5 ) m g = + 1 m g \begin{aligned} F_B &= - m \frac{v_B^2}{R} - mg = - \left(\frac{2 h}{R} + 1 \right) m g = - 7 \cdot m g \\ F_C &= + m \frac{v_C^2}{R} - mg = + \left(\frac{2 h}{R} - 5 \right) m g = + 1 \cdot m g \\ \end{aligned}

Ya. It would have been trickier if the velocity turned up to be zero at any point in the circular groove. Then the roller coaster will leave track and follow a projectile (Until every passenger dies) :P

Md Zuhair - 3 years, 9 months ago

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