Question

A 1.50 kilogram cart is traveling in a horizontal circle with a radius of 2.4 meters at constant speed of 4.00 meters per second. • Calculate the minimum coefficient of static friction between the wheels of the cart and the surface. Show all work, including equations, substitutions and units. • Calculate the time of one complete revolution. • Describe a change that would quadruple the magnitude of the centripetal force. By what factor would the coefficient of friction change to accommodate this new force?

Answer 1

the required centripetal force is given by the static frictional Force

centripetal force = static frictional Force

m*v^2/R = mu_s*N

m*v^2/R = mu_s*m*g

m cancels on both sides

v^2/R = mu_s*g

mu_s = v^2/(R*g)

mu_s = (4^2/(2.4*9.81)) = 0.679 no units

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time period is T = (2*pi*r)/v = (2*3.142*2.4)/4 = 3.77 sec

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centripetal Force is F =m*v^2/R

to quadraple the force ,we need to increase the speed by twice

new coefficient of static friction is mu_s' = v'^2*/(r*g) = 4*mu_s = 4*0.679 = 2.716

required factor is (2.716-0.679)/0.679 =3