A mass m=76kg slides on a frictionless track that has a drop, followed by a loop

Question

A mass m=76kg slides on a frictionless track that has a drop, followed by a loop-the -loop with radius R=19.9m and finally a flat straight section at the same height as the center of the loop (19.9m off the ground). Since the mass would not make it round 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.) What is the minimum speed the block must have at the top to make it round the loop-the-loop with out leaving the track m/s and the height above the ground must the mass begin to make it round the loop-the-loop m. So the mass has just enough speed to make it round the loop without leaving the track, what will its speed be at the loop m/s and its speed at the final flat level(19.9m off the ground) m/s.

Explanation / Answer

At the top of the loop. The force acting on the car is mg-Normal force from the track. The sum of the forces must equal mv^2/R to maintain the car's circular motion. Assuming Normal force from the track is equal to zero (or a very small negliglable amount, then mg must provide the centripetal force to keep the car in circular motion. mg = mv^2/R g = v^2/R v = sqrt (gR) = sqrt(9.81*19.9) = 13.97 m/s Final velocity can be found by conserving energy. 0.5*m*vfinal^2 = 0.5*m*v^2 (from the top of the loop) + m*g*R vfinal^2 = v^2 + 2*g*R vfinal = sqrt(v^2 + 2*g*R) = sqrt(13.97^2 + 2 * 9.81 * 19.9) = 24.2 m/s

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