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https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/summer/homework/Ch-07-GPE-ME/loop_the_loop_spring/2.gif

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.)

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

Explanation / Answer

At top of loop Mv2/r =N + mg = 1.8 mg So v=14.55 m/s So at height of R Using energy conservation MgR + 1/2 mv2 = 1/2 m vf^2 So vf = 21.14 m/s

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