The force of gravity and the normal force acting on a passenger in the roller co

Question

The force of gravity and the normal force acting on a passenger in the roller coaster at the top of the loop are shown in following figure (a). The sum of these two forces is the net force, which has to equal the centripetal force in circular motion. If the net force (centripetal force here) is equal to the gravitational force, then the normal force is zero, and the passenger feels weightless. This situation is illustrated in following figure (b). We just stated that the net force is equal to the centripetal force and that the net force is the sum of the normal force and the force of gravity: For the feeling of weightlessness at the top of the loop, we need N = 0, and thus: F_c = F-g. Roller Coaster problem, suppose the vertical loop has a radius of 6m and the passenger mass is 100.0 kg. Calculate the liner speed of 3 o'clock and normal force at 6 o'clock.

Explanation / Answer

(a) it is given that centripetal force= force due to gravity

mv2 / r =m*g

where m is mass of body

r is radius

v is linear speed

g is acceleration due to gravity

v=sqrt g*r

=sqrt 9.8 *6

linear speed at 3 o clock is v=7.668 m/sec

(b)linear speed at 6 o clock=sqrt 9.8*12=10.844 m/sec

(c)normal force N=m*g=100*9.8=980 N

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