normal for and minimum kinetic energy of a loop?
(part 1)A 3.8 g mass is released from rest at C which
has a height of 1.1 m above the base of a
loop-the-loop and a radius of 0.26 m.
The acceleration of gravity is 9:8 m=s2 :
h = 1.1 m
radius of loop = 0.26 m
B = top point of loop
D = bottom point of loop
C = release point
A = half way up loop at 3oclock (half way between B and D
mass = 3.8 g
Find the normal force pressing on the track at A, where A is at the same level as the center of the loop. Answer in units of N.
(part 2) Consider a different situation when the initial height at C has not yet been specified.
What is the minimum kinetic energy of the block at B, which is located at the top of the loop, so that the block can pass by this point without falling off from the track? Answer in units of J.
During its fall the body accepts only the gravity force on it. Gravity is a conservational force, so we can use the principle of conservation of mechanical energy on the mass. At C the mass has a potential energy equal to mgh (assuming this energy is zero at the base of the loop) or 3.8 kg x 9.8 ms^-2 x 1.1 m.
At A, all this energy has been transformed into kinetic energy, or mu^2/2. You solve for u (or u^2, it doesn't matter). Again at A, the body has entered the loop and performs a circular motion. By necessity the force from the loop (normal one) acts as a centripetal force and is equal to mu^2/r, where r is the radius of the loop. At B, the centripetal acceleration should equal gravity, g=u^2 and min. K =mu^2/2 (enough u for the mass to be able to ignore the gravity force as a downwards-pulling force and participate in the circular motion, this is how you could interpret it).
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US $799.00
