A roller coaster of mass m = 1500. kg is on a certain track (Fig. 1). The kineti

Question

A roller coaster of mass m = 1500. kg is on a certain track (Fig. 1). The kinetic coefficient of friction between the roller coaster and the track is mu k = .30 along the segment AB. All other segments are frictionless. Assume m starts from rest at the top of the hill. Determine the velocity vB at point B. Determine the maximum radius R of the loop such that m is on the verge of losing contact with the track at point C. Use this R for the remainder of the problem. Determine the velocity vD at point D. Determine the horizontal distance between points D and E

Explanation / Answer

intial energy of system = m * g * h = 1500 * 9.8 * 182 = 2,675,400 Joules

friction force = mu * normal reaction = 0.3 * m*g*cos 78 = 916.89056 N ..

energy dissipated by friction = work done by friction = friction * distance

so energy dissipated by friciotn = 916.89056 * 84 = 77,018.81 joules ...

so final energy of block at point B = initial energy - energy dissipated

= 2,675,400 - 77,018.81 = 2,598,381.19

This final energy = potential energy + kinetic

Let the velocity at B be vb

height hB = 91 meters

so 2,598,381.19 = m*g*hB + 0.5 * m * vB^2

2,598,381.19 = 1500*9.8 * 91 + 0.5 * 1500 * vB^2

so vB = 40.9989 m/sec = 41 m/sec ( approx )

b) Let the radius be R ....

so at C as it is just loosing contact..

so m*vC^2 / R = mg .... no other force will be there

so vC = sqrt ( g*R )

energy at point B = 2,598,381.19

hC = hB + 2R = 91 + 2R

energy at point C = m*g*(hC) + 0.5 * m* vC^2

= 1500*9.8*(91 + 2R ) + 0.5 * 1500 * g*R

conserving energy

1500*9.8*(91 + 2R ) + 0.5 * 1500 * 9.8*R = 2,598,381.19

so.. 36,750 * R = 1,260,681.19

so R = 34.30425 meters ...

c)

Energy at point B = 2,598,381.19

Let the velocity at point D be vD

heigth hD = R - R cos 45 = 34.30425 * ( 1 - cos 45 ) = 10.0475

so energy at point D = m*g*(hD) + 0.5 * m* vD^2 = 1500*9.8*10.0475 + 0.5 * 1500 * vD^2

conserving energy

147,697.9884 + 0.5 * 1500 * vD^2 = 2,598,381.19

so vD = 57.16273 m/sec

d)

for the projectile ...

along vertical ...

intial velocity u = vD sin 45 = 57.16273 * sin 45 = 40.420153 m/sec

distance travelled = s = - hD = - 10.0475 meters .. ( -ve because it travels along -ve y axis )

acceleration a = -9.8 m/sec2

let the time taken be t ...

s = ut + 0.5 * a*t^2

-10.0475 = 40.420153 - 0.5 * 9.8 * t^2

4.9t^2 - 40.420153 t - 10.0475 = 0

so time take t = 8.49052 secs....

along horizontal ....

velocity = vD cos 45 = 57.16273 * cos 45 = 40.420153 m/sec

so horizontal distance = velocity * time .... as there is no acceleration along horizontal direction

so horozontal distance = 40.420153 * 8.49052 = 343.188 m

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