A mass m = 89.0 kg slides on a frictionless track that has a drop, followed by a

Question

A mass m = 89.0 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.9 m and finally a flat straight section at the same height as the center of the loop (15.9 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 in m/s? AND What height above the ground must the mass begin to make it around the loop-the-loop in m?

Explanation / Answer

the centrifugal force must just cancel the gravity force. m*v^2/r = m*g v^2/r = g = 9.8m/s^2 v^2 = 17.9*9.8 = 175.42 v = 13.24 m/s how far would a block have to fall to gain this speed? position = X = X0 + V0*T + 1/2A*T^2 X0 = 0, V0 = 0 X = 0.5*a*T^2 velocity = dX/dT = at =9.8*T 13.24/9.8 = T = 1.35 sec avg speed = Vmax/2 therefore distance = 6.62m/s * 1.35 sec = 8.95M the block must start 8.95M above the top of the loop or 35.8 +8.95 = 44.75M above the ground the block will have 44.75M * 73kg*9.8M/s^2 = 32.014KJ of kinetic energy K.E. = 0.5*m*V^2 = 0.5 * 73 * v^2 = 32.014KJ V^2 = 877.1 V = 29.62M/s at the final flat, we have to subtract the potential energy of a 17.9M rise from its' kinetic energy 32014 - 17.9*73*9.8 = 19208J 0.5*73*V^2 = 19208 V^2 = 22.94M/s the potential energy in a spring is 0.5 * K * X^2 = 19208J X^2 = 19208/0.5K = 2.48 in^2 X = 1.57 in

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