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The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 12 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.8 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

c) When the car is descending vertically (ie at a height R above the ground) in the loop, what is its speed |v|?

d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 20 m. What retarding acceleration |a| is required?

Explanation / Answer

a)Conserving energy between top of the loop and the point of release,

mgh=0.5mv2

v2=2gh

Balancing forces at the top of the loop,

mv2/r-mg=0.8mg

Solving, v= sqrt(1.8gR)=14.55m/s

h=v2/2g=10.8m

b)Let the speed be u.

Conserving energy,

0.5mv2+mgR=0.5mu2

Solving, u=21.14m/s

c)Velocity at the bottom of the loop=26.12m/s

Let the deceleration be a.

0-26.122=2a*20

a=-17.05m/s2

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