A 1.50 kilogram cart is traveling in a horizontal circle with a radius of 2.4 me
ID: 1876104 • Letter: A
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?
Explanation / Answer
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
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