Question

Six roller-coaster carts pass over the same semicircular "bump." (Figure 1 ) The mass M of each cart (including passenger) and the normal force n of the track on the cart at the top of each bump are given in the figures.

Rank the speeds of the different carts search passes over the top of the bump.

Rank from largest to smallest. To rank items as equivalent, overlap them.

a) 200N, 400kg

b) 400N, 100kg

c) 300N, 300kg

d) 800N, 100kg

e) 800N, 800kg

f) 400N, 200kg  

Answer 1

Concepts and reason

The concepts used to solve this question are centripetal force and Newton's second law of motion, mass, acceleration radius, and radius of the circle. Initially, use Newton's second law of motion to calculate the net force acting on the cart at the top of the semicircular bump and use the expression for the centripetal force to calculate the speed of the cart at the top of the bump.

Fundamentals

According to Newton's second law of motion, the net force on an object is equal to the product of mass \(\mathrm{m}\) and acceleration

a. The expression for the force is, \(F_{\text {net }}=m a\)

Here, \(m\) is the combined mass of the cart and passenger. The expression for the centripetal force on a circularly rotating object is as follows:

\(F_{\mathrm{C}}=\frac{m v^{2}}{r}\)

Here, \(v\) is the speed of the object and \(r\) is radius of the circle.

\(\mathrm{n}\) is the normal force, \(\mathrm{M}\) is the mass of roller coaster cart and passenger, \(\mathrm{g}\) is the acceleration due to gravity, and \(\mathrm{r}\) is the radius of circle. From the free body diagram, the net force on the roller coaster car when it is at top of the circle is given as follows:

\(F_{\mathrm{net}}=M g-n\)

Since the roller coaster moving in a circle, the net force on the roller coaster must be equal to the centripetal force. \(F_{\mathrm{net}}=\frac{M v^{2}}{r}\)

Substitute \(\frac{M v^{2}}{r}\) for \(F_{\text {net }}\) in the above equation \(F_{\text {net }}=M g-n\) \(\frac{M v^{2}}{r}=M g-n\)

Rearrange the above equation for speed \(v\). \(v=\sqrt{\left(g-\frac{n}{M}\right) r}\)

The net force on the roller coaster cart is equal to the centripetal force \(\frac{m v^{2}}{r}\) is always directed towards the center of the circle, \(\mathrm{N}\) is normal force directs upwards, and \(\mathrm{mg}\) is weight of roller coaster car directs downwards. The derived equation for the speed of the roller coaster cart depends on the acceleration due to gravity, normal force, the radius of the semicircular bump and the combined mass of the cart and passenger.

The expression for the speed of roller coaster cart (a) is, \(v_{\mathrm{a}}=\sqrt{\left(g-\frac{n}{M}\right) r}\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}\), \(200 \mathrm{~N}\) for \(\mathrm{n}\), and \(400 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation.

$$ \begin{array}{c} v_{\mathrm{a}}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{200 \mathrm{~N}}{400 \mathrm{~kg}}\right) r} \\ =\sqrt{9.3 r} \end{array} $$

\(=(3.04) r\)

The expression for speed of roller coaster cart (b) is, \(v_{\mathrm{b}}=\sqrt{\left(g-\frac{n}{M}\right) r}\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}, 400 \mathrm{~N}\) for \(\mathrm{n}\), and \(100 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation and solve for \(\mathrm{v}\) in terms of \(\mathrm{r}\)

$$ \begin{array}{c} v_{\mathrm{b}}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{400 \mathrm{~N}}{100 \mathrm{~kg}}\right) r} \\ =\sqrt{5.8 r} \\ =(2.4) r \end{array} $$

The expression for speed of roller coaster cart (c) is \(v_{\mathrm{c}}=\sqrt{\left(g-\frac{n}{M}\right) r}\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}\), \(300 \mathrm{~N}\) for \(\mathrm{n}\), and \(300 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation and solve for \(\mathrm{v}\) in terms of \(\mathrm{r}\)

\(v_{c}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{300 \mathrm{~N}}{300 \mathrm{~kg}}\right) r}\)

$$ \begin{array}{l} =\sqrt{8.8 r} \\ =(2.96) r \end{array} $$

The expression for speed of roller coaster cart (d) is, \(v_{\mathrm{d}}=\sqrt{\left(g-\frac{n}{M}\right) r}\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}, 800 \mathrm{~N}\) for \(\mathrm{n}\), and \(100 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation and solve for \(\mathrm{v}\) in terms of \(\mathrm{r}\)

$$ \begin{array}{c} v_{d}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{800 \mathrm{~N}}{100 \mathrm{~kg}}\right) r} \\ =\sqrt{1.8 r} \\ =(1.34) r \end{array} $$

The expression for speed of roller coaster cart (e) is,

$$ v_{\mathrm{e}}=\sqrt{\left(g-\frac{n}{M}\right) r} $$

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}, 800 \mathrm{~N}\) for \(\mathrm{n}\), and \(800 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation and solve for \(\mathrm{v}\) in terms of \(\mathrm{r}\).

$$ \begin{array}{c} v_{e}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{800 \mathrm{~N}}{800 \mathrm{~kg}}\right) r} \\ =\sqrt{8.8 r} \\ =(2.96) r \end{array} $$

The expression for speed of roller coaster cart (f) is,

$$ v_{f}=\sqrt{\left(g-\frac{n}{M}\right) r} $$

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}, 800 \mathrm{~N}\) for \(\mathrm{N}\), and \(800 \mathrm{~kg}\) for \(\mathrm{M}\) in the above equation and solve for \(\mathrm{v}\) in terms of \(\mathrm{r}\)

$$ \begin{array}{c} v_{f}=\sqrt{\left(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)-\frac{400 \mathrm{~N}}{200 \mathrm{~kg}}\right) r} \\ =\sqrt{7.8 r} \\ =(2.79) r \end{array} $$

The speeds of the all roller coaster carts are calculated by using the equation for the speed of the roller coaster that derived in the step \(1 .\) The speed of the roller coaster depends on the acceleration due to gravity, normal force, mass and the radius of the semicircular bump.

The speeds of the roller coaster carts are calculated in step2. The ranking of the speeds can be decided by depending on the values of the speeds calculated in the step2. The ranking of the speeds for the roller coaster carts from largest to smallest is, \(a>c=e>f>b>d\)

The rank of speed of roller cart from largest to smallest is \(\mathrm{a}>\mathrm{c}=\mathrm{e}>\mathrm{f}>\mathrm{b}>\mathrm{d}\).

The speeds of the all roller coaster carts are calculated by using the equation for the speed of the roller coaster that derived in the step \(1 .\) Depending on the values of the speeds of the roller coasters in step 2 , the ranking of the speeds of the roller coaster is decided.

Answer 2

At the top,

Net force on the cart, mg -N provides the centroipetal force, mv2/R

Hence,   mv2/R = mg - N

             v2/R = g - N/m = (9.8-N/m; )

     So the ranking can be done onthe basis of the values of(g-N/m)           

Calculate N/m values for each

    N     m          N/m  ( g -N/m)   Rank      

a) 200N, 400kg ;  0.5   9.3          1    

b) 400N, 100kg   4.0    5.8          4

c) 300N, 300kg     1.0    8.8          2

d) 800N, 100kg     8.0     1.8          5

e) 800N, 800kg     1.0     8.8          2

f) 400N,200kg      2.0      7.8          3

The ranking intheorder of decreasingspeed(largest first)

   a, c&e, f, b, d