The small mass m sliding without friction along the looped track shown in Fig. 1

Question

The small mass m sliding without friction along the looped track shown in Fig. 1 is to remain on the track at all times, even at the very top of the loop of radius R. In terms of the given quantities, determine the minimum release height for the mass to stay on the track. If the actual release height is 2h, calculate the normal force at: the track at the bottom of the loop, The track at the top of the loop, and by the track after the block exits the loop and is now on the flat surface. When the mass enters the flat surface, there is a coefficient of kinetic friction Mu_k which slows the mass to a stop. What distance will the mass travel before coming to rest?

Explanation / Answer

a) In order for the mass to remain on the track at the top of the loop, the centrifugal force there must at least equal the weight of the mass

CF = m*v²/r
Wt = m*g

Condition to stay on track CF > Wt; m*v²/r = m*g; v > [r*g]

The kinetic energy at the top of the loop is then ½*m*v² = ½*m*r*g

The potential energy change from the release point to the top of the loop PE = m*g*y
y = h - 2*r; PE = m*g*(h - 2*r)

Equate KE to PE to get 1/2*m*r*g = m*g*(h - 2*r)
1/2*r = h - 2*r
h > (1/2*r + 2*r) = 2.5*r

b)    Given H = 2h = 5r

At the bottom of the loop
PE = m*g*H = m*g*5r
KE = 1/2*m*v²
v² = 10*g*r
CF = m*v²/r = 10*m*g
Weight = m*g
Total force against track at bottom = 11*m*g

c)   At the top of the loop
PE = m*g*(H - 2*r) = m*g*(5r - 2*r) = m*g*3r
KE = 1/2*m*v²
m*g*3r = 1/2*m*v²
v² = 6.0*g*r
CF = m*v²/r = 6.0*m*g
normal force = CF - m*g = 5.0*m*g

d) When the block is on the flat section, the normal force is the weight   = m*g

e)    distance travelled =    mg * 2r / (kmg + mg)

                                      = 2r / (k + 1 )

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