A student of mass M = 66 kg takes a ride in a frictionless loop-the-loop at an a

Question

A student of mass M = 66 kg takes a ride in a frictionless loop-the-loop at an amusement park. The radius of the loop-the-loop is R = 13 m. The force due to the seat on the student at the top of the loop-the-loop is FN = 544 N and is vertically down. What is the apparent weight of the student at the bottom of the loop-the-loop?

Have tried this problem multiple times, the aswers ive gotten are: 749.6, 1658.8, 1190.6, 5608.7, 1839, and 3336.8. All have been wrong.

My professor said you need to consider that the speed will change as the roller coaster moves around the track and to use energy conservation to find the speed at the bottom.

Explanation / Answer

M = 66 kg R = 13 m   Fn = 544 N

When at the top the seat exerts a force Fn = 544 N

Weight of the boy = 66 x 10 = 660 N

This the force acting towards the center of the circle or centripetal force

Fcp = MV2 /R = 544 N+ 660 N = 1204 N

KE at the top = MV2/2 = R xFcp /2 = 13 x 1204 /2= 7826 J

change in PE from top to bottom = M x g x 2R

                                                    = 66 x 10 x 26 we take g = 10 for simplicity

KE at the bottom = 7826 + 17160 = 24986 J

Force toward the center at the bottom = Fcp = 2 x KE /R = 2 x 24986 /13 = 3844 N

The real weight of the boy Mg = 660 N act down wards

Apparent weight = 660 - 3844 = -3184 N

The radius R seems to be given wrong hene we are seeing a -ve apparent weight, R may be 1.3 M instead of 13M, please check

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