Loop the Loop (w/Work) A mass m 87 kg slides on a frictionless track that has a

Question

Loop the Loop (w/Work) A mass m 87 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R 15.3 m and finally a flat straight section at the same height as the center of the loop (15.3 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? 12.244 2) What height above the ground must the mass begin to make it around the loop-the-loop? 38.250 m Submit Help 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?

Explanation / Answer

m = 87 Kg

R = 15.3 m

1) at the top of the loop

let the minimum velocity is v

m * v^2/R = m * g

v^2/15.3 = 9.8

v = 12.2 m/s

for the parts 5

5) let the compression is x

Using conservation of energy

0.50 * 15900 * x^2 = 0.50 * 87 * 21.2^2

solving for x

x = 1.57 m

the spring will be compress 1.57 m

6)

let the initial velocity needed is u

Using conservation of energy

initial mechanical energy = final mechanical energy

0.50 * m * u^2 + m *g * 2R = 0.50 * m * 12.2^2 + m *g * 2R

solving for u

u = 12.2 m/s

the speed needed for initial push is 12.2 m/s

7)

as normal force is always perpendicular to velocity ,

the work done by normal will be zero

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