Question

A \({\bf{350}}\;{\bf{kg}}\) roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Fig. P7.41). (a) How fast is this roller coaster car moving at point B? (b) How hard does it press against the track at point B?

Short answer

a.) The speed of the roller coaster car at point B is \(15.96\;{\rm{m/s}}\)

b.) the force exerted on the track by the roller coaster at point B is \(11433.3\;{\rm{N}}\)

Step-by-step solution

Identification of given data

The given data can be listed below,

• Mass of the roller coaster is\({\rm{350}}\;{\rm{kg}}\)
• The height at point A is\({\rm{25}}{\rm{.0}}\;{\rm{m}}\)
• The height at point B is \({\rm{12}}{\rm{.0}}\;{\rm{m}}\)

Concept/Significance of conservation law

According to conservation law, a conserved quantity for a system that goes through a process is constant both before and after the process.

(a) Determination of the speed of the roller coaster car at point B

The roller coaster is at rest at the initial position. So, the initial kinetic energy \({K_i}\) is zero.

The gravitational potential energy is expressed as,

\({U_{grav}} = mgy\)

The gravitational potential energy at point A is obtained by substituting the given values for A, in the above equation,

\(\begin{aligned}{}{U_A} &= \left( {350\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {25.0\;{\rm{m}}} \right)\\ &= 85750\;{\rm{J}}\end{aligned}\)

Similarly, thegravitational potential energy at point Bis obtained by substituting the given values for B, in the above equation,

\(\begin{aligned}{}{U_B} &= \left( {350\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {12.0\;{\rm{m}}} \right)\\ &= 41160\;{\rm{J}}\end{aligned}\)

According to law of conservation of energy, the decrease in gravitational potential energy should be equal to increase in kinetic energy. As the kinetic energy is zero at A, so equation becomes-

\(K = {U_A} - {U_B}\)

For \({U_A} = 85750\;{\rm{J}}\) and \({U_B} = 41160\;{\rm{J}}\), the kinetic energy of the roller coaster at B is given as-

\(\begin{aligned}{}K = 85750\;{\rm{J}} - 41160\;{\rm{J}}\\ = 44590\;{\rm{J}}\end{aligned}\)

Now, the final kinetic energy is expressed as,

\(K = \frac{1}{2}m{v^2}\)

For,\(K = 44590\;{\rm{J}}\)and \(m = 350\;{\rm{kg}}\), the velocity of the roller coaster at B is -

\(\begin{aligned}{}44590 &= \frac{1}{2}\left( {350\;{\rm{kg}}} \right)\left( {{v^2}} \right)\\v &= \sqrt {\frac{{2 \times 44590\;{\rm{J}}}}{{350\;{\rm{kg}}}}} \\v &= 15.96\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the roller coaster is \(15.96\;{\rm{m/s}}\).

(b) Determination of the force exerted on point B.

The force with which the tracks are pressed at the point B should be equal to the normal reaction force\(\left( n \right)\)by the tracks, if\(a\)is the centrifugal acceleration of the roller coaster and\(g\)is the acceleration due to gravity of earth, then-

\(\begin{aligned}{}ma &= n + mg\\n &= m\left( {a - g} \right)\end{aligned}\)

Substitute the value of acceleration in the above equation.

\(\begin{aligned}{}n& = m\left( {\frac{{{v^2}}}{R} - g} \right)\\ &= 350\;{\rm{kg}}\left( {\frac{{{{\left( {15.96\;{\rm{m/}}{{\rm{s}}^2}} \right)}^2}}}{{6\;{\rm{m}}}} - 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ &= 11433.3\;{\rm{N}}\end{aligned}\)

Thus, the force exerted on the track at point B is \(11433.3\;{\rm{N}}\)