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

Question

A mass m 76 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R 19.6 m and finally a flat straight section at the same height as the center of the loop (19.6 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 Submit 2) What height above the ground must the mass begin to make it around the loop-the-loop? m Submit 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 Submit 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 (19.6 m off the ground)? m/s Submit 5) Now a spring with spring constant k 17100 N/m is used on the final flat surface to stop the mass. How far does the spring compress? m Submit 6) It turns out the engineers designing the loop-the-loop didn't really know physics hen they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning. How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track? m/s submit

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 = 19.6*9.8
v = 13.86 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
13.86/9.8 = T = 1.41 sec
avg speed = Vmax/2 therefore distance = 6.93m/s * 1.41 sec = 9.77M
the block must start 9.77M above the top of the loop or 39.2 +9.77 = 48.97M above the ground

the block will have 48.97M * 76kg*9.8M/s^2 = 36.472KJ of kinetic energy
K.E. = 0.5*m*V^2 = 0.5 * 76 * v^2 = 36.472KJ

V = 30.98M/s

at the final flat, we have to subtract the potential energy of a 19.6M rise from its' kinetic energy
36472 - 19.6*76*9.8 = 21873J
0.5*76*V^2 = 21873 V^2 = 23.99M/s

the potential energy in a spring is 0.5 * K * X^2 = 21873J
X^2 = 21873/0.5K = 2.48
X = 1.59

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