A car rolls without friction around the loop track in an amusement park. It star

Question

A car rolls without friction around the loop track in an amusement park. It starts from rest at point

A, height h above the bottom of the loop. A few gran

A car rolls without friction around the loop track in an amusement park. It starts from rest at point A, height h above the bottom of the loop. A few gran' mammas, who ditched Sunday School for some better thrills, embark on the joy ride What is the minimum value of h for the car to ride around the loop without falling off at the top (point B). Express this value in terms of R We now assume h = (7 2)R What is the reaction force of the track at point B? Compute the speed, radial and tangential accelerations when the car is passing point C (mid point). Show these acceleration components in a vector diagram drawn to scale How many g's are the thrilled gan' mammas experiencing at the bottom of the loop?

Explanation / Answer

let vo is the velocity of the car at the bottom

0.5*m*vo^2 = m*g*h

vo = sqrt(2*g*h)


the minimum speed that car sgould have at the top point

v = sqrt(g*R)


total mechanical enrgy at the bottom = mechanical energy at the top point

0.5*m*vo^2 = 0.5*m*v^2 + m*g*(2*R)

0.5*m*2*g*h = 0.5*m*g*R + 2*m*g*R

m*g*h = 0.5*m*g*R + 2*m*g*R

h = 2.5*R

b)

h = (7/2)*R

m*g*h = 0.5*m*v^2 + 2*m*g*R


m*g*3.5*R = 0.5*m*v^2 + 2*m*g*R

0.5*m*v^2 = 1.5*m*g*R

vB = sqrt(3*g*R)


i) N = mg - m*v^2/R

= m*g - m*3*g*R/R

= -2*m*g

ii)

0.5*m*Vc^2 = m*g*R + 0.5*m*VB^2

0.5*m*VC^2 = m*g*R + 0.5*m*3*g*R

0.5*Vc^2 = g*R + 1.5*g*R

Vc = sqrt(5*g*R)

a_tan = g = 9.8 m/s^2

a_rad = VC^2/R = 5*g*R/R = 5*g

iii)

at the bootom

V = sqrt(2*g*3.5*R)

v = sqrt(7*g*R)

a(bottom) = g + v^2/R

= g + 7*g

= 8*g

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