C1991M1. A small block of mass 3m moving at speed v_0/3 enters the bottom of the

Question

C1991M1. A small block of mass 3m moving at speed v_0/3 enters the bottom of the circular, vertical loop-the-loop shown above, which has a radius r. The surface contact between the block and the loop is frictionless. Determine each of the following in terms of m, v_0, r, and g. The kinetic energy of the block and bullet when they reach point P on the loop The speed V_min of the block at the top of the loop to remain in contact with track at all times The new required entry speed v_0' at the bottom of the loop such that the conditions in part b apply.

Explanation / Answer

a)

at point P :

h = r/2

KEp = kinetic energy at P

KEi = initial kinetic energy = (0.5) (3m) (vo/3)2 = 0.167 m vo2

using conservation of energy

KEi = KEp + PE

0.167 m vo2 = KEp + mgh

0.167 m vo2 = KEp + mg (r)

KEp = 0.167 m vo2 - mgr      

b)

at the top of loop , force equation is given as

Fn + mg = m V2min/r

for losing the contact

Fn = 0

hence

mg = m V2min/r

Vmin = sqrt(gr)

c)

using conservation of energy

KEi = KEt + PEt

(0.5) m v'2 = (0.5) m v2min + m g(2r)

v'2 = (sqrt(gr))2 + g(4r)

v' = sqrt(5gr)

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