3. A particle with mass m starts from point A and goes along track ABC according

Question

3. A particle with mass m starts from point A and goes along track ABC according to the figure. The plane of the track is vertical, segment AB is smooth while BC is rough. The radius R of segment AB and other data are given belovw Data R-301771], =60 7n VC =0 Questions a, Calculate the magnitude of the reaction force that acts on the particle in point A b, Write up the work-energy theorem on segment AB and calculate the magnitude of the velocity of the particle in point B c, Write up the work-energy theorem on segment BC and calculate distance As between points B and C. (The particle stops in point C) d, Calculate the scalar acceleration of the particle on segment BC. e, Calculate the magnitude of the reaction force that acts on the particle on segment BC

Explanation / Answer

a. let the reaction force at point A be N

then from force balance

N = mgcos(alpha) + mv^2/R

now given

alpha = 60 deg

v = 15 m/s

R = 30 m

hence

N = m(9.81*cos(60) + 15^2/30) = 12.405m N ( where m is mass of the object)

b. as the path between AB is friction less, total energy of the mass is conserved

so initial total energy = Potential + kinetic energy = 0.5mv^2 + mgR(1 - cos(alpha)) = m(0.5*15^2 + 9.81*30(1 - cos(60))) = 259.65m

final total energy = 0.5mu^2 ( where u is final speed at B)

hence from work energy theorem

259.65m = 0.5mu^2

u = 22.788 m/s

c. the friction force acting in segment BC is k*mg ( where k = 0.2)

sop let the distance BC be s

then

k*mg*s = 259.65m ( as final speed of the block is 0, it has 0 J of final energy)

0.2*9.81*m*s = 259.65m

hence

s = 132.339 m

d. for constant acceleration a

2*a*s = u^2

2*a*132.339 = 22.788^2

a = 1.96 m/s/s

e. reaction force on the particel in segment BC = mg = 9.81m ( where m is mass of the object)

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