The hoop thin ring has a mass of 5 kg and is released down t

The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin omega = 9 rad/s and its center has a velocity nu g = 5 m/s as shown. The coefficient of kinetic friction between the hoop and the plane is mu_k = 0.4. (Figure 1) Determine how long the hoop rolls before it stops slipping. Express your answer to three significant figures and include the appropriate units. t = 3.59 s

Solution

Weight W = mg = 5*9.81 = 49.05 N

Normal force on incline N = W*cos30 = 49.05*cos30 = 42.4785 N

Friction force F = 0.4 * 42.4785 = 16.9914 N

Force along the incline = W*sin30 = 49.05 * sin30 = 24.525 N

MOI of hoop = mr² = 5*0.5² = 1.25 kg-m²

By linear momentum conservation,

m*v_G + Ft = m*v_Gnew

5*5 + 24.525*t - 16.9914*t = 5*v_Gnew..............eqn1

By angular momentum conservation,

-1.25*9 + 16.9914*0.5*t = 1.25*(v_Gnew / 0.5)...................eqn2

Solving eqn1 and eqn2 we get

t = 5.022 s

vGnew = 12.567 m/s