A mass m = 86.0 kg slides on a frictionless track that has a drop, followed by a

Question

A mass m = 86.0 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.5 m and finally a flat straight section at the same height as the center of the loop (19.5 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.) 1) What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? m/s 2) What height above the ground must the mass begin to make it around the loop-the-loop? m 3) If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? m/s 4) If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (19.5 m off the ground)? m/s 5) Now a spring with spring constant k = 18900.0 N/m is used on the final flat surface to stop the mass. How far does the spring compress? m 6) It turns out the engineers designing the loop-the-loop didn

Explanation / Answer

i) If the mass starts from rest at the same height "2R" as the top of the loop then when it reaches the top of the loop it again must come to rest, otherwise it will have more energy then it started with. This is only for a moment, as the mass would then start from rest and drop straight down. To find the minimum velocity at the top of the loop you have to do a force analysis. At the top, if the mass is moving with velocity "v" there are ,in general, two forces on the block; 1)its weight "mg" and 2)the normal force "N" of the track. Both forces act down and their net is the centripetal force which obeys the force eq; N + mg = mv^2/R The minimum velocity allowed, and still allow circular motion, will occur when the normal force just equals zero. The mass is barely in contact with the track. This gives you the velocity asked for; 0 + mg = mv^2/R v^2 = gR v = sqrt(gR) = sqrt(9.8*19.5) = 13.82 m/s 2) To find the height "h" needed to give this velocity use conservation of energy. Initial PE = final PE + KE; mgh = mg(2R) + (1/2)mv^2 h = 2 R + 1/2v^2/g = 2*19.5 + 1/2 * 13.82^2* / 9.8 = 48.74 m 3) To find the velocity at the bottom (0 PE) just apply conservation of energy again; mgh = 0 + (1/2)mv^2 v^2 = 2gh = 5gR Energy: ½mVt² + m*g*(2*R) = ½m*Vb² ? Vb² = Vt² + 4*g*R Vb = 30.91 m/s 4) On the level again apply conservation of energy; mgh = mgR + (1/2)mv^2 g(5R/2) = gR + (1/2)v^2 v^2 = 3gR Vf² = Vt² + 2*g*R Vf² = 13.82² + 2*9.8*19.5 ? Vf = 23.94 m/s 5) Es = KE ½kx² = ½mVf² ? x² = m*Vf²/k x² = 86*23.94²/18900 x = 1.614 m 6) At top, Ac = g = v²/R ? Vtmin = v(Rg) = sqrt(19.5*9.8)= 13.82 m/s

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