Question

A 0.442 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius of 1.27 m. The particle is given an initial speed of 8.15 m/s. After one revolution, its speed has dropped to 6.11 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.Calculate the coefficient of kinetic friction.What is the total number of revolutions the particle makes before stopping? Do not enter units.

Answer 1

By energy conservation law,

energy loss due to friction(E) = -(loss in kinetic energy(dKE))

E = -(0.5*m*vf^2 - 0.5*m*vi^2)

here, vf = final speed = 6.11 m/sec.

vi = initial speed = 8.15 m/sec.

m = mass of particle = 0.442 kg

So, E = -0.5*0.442*(6.11^2 - 0.442^2)

E = 8.21 J

Since E = fr*d

here, fr = friction force = $\mu$k*m*g

d = distance travel by ball = n*2*pi*r

where, n = revolution = 1

r = 1.27 m

So, 8.21 = $\mu$k*0.442*9.81*1*2*pi*1.27

$\mu$k = 8.24/(0.442*9.81*2*pi*1.27)

$\mu$k = 0.238

Let ball comes to rest after N revolution,

then, KEf = 0

energy loss due to friction(E) = -(KEf - KEi)

E = KEi = 0.5*m*vi^2 = 0.5*0.442*8.15^2

E = 14.679 J = fr*d'

here d' = N*2*pi*r

So, $\mu$k*m*g*N*2*pi*r = 14.679

N= 14.679/(0.238*0.442*9.81*2*pi*1.27)

N = 1.782 rev

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