How fast is the wheel moving when it reaches the foot of the hill if it rolled w

Question

How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down?

How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

Explanation / Answer

Wheel will have both translation energy and rotational energe KE total(initial)= 0.5*m*v^2 +0.5*I*w^2 [ v is velocity of center of mass, w is angular velocity and I is moment of inertia] w=v/r= 11/0.85 =12.9411 rad/sec I= 0.5* m*r^2 [ r is radius of cylinder] KE final = KE initial +mgh [h=75] =>0.5*m*vf^2 +0.5*I*wf^2 = 0.5*m*v^2 +0.5*I*w^2 +mgh [ vf and wf are final velocity and angular velocity respectively] =>0.5*m*vf^2 + 0.25*m*r^2*vf^2/r^2 = 0.5*m*v^2 +0.25*m*r^2*v'^2/r^2 +mgh => 0.5*vf^2 + 0.25*vf^2 = 0.5*v^2 + 0.25*v^2 + gh =>0.75 vf^2=0.75 v^2+gh [v=11 m/sec, h=75m] =>vf=33.181 m/sec b)KE total = 0.5*m*vf^2 +0.5*I*wf^2 [I=0.5*m*r^2 , wf = vf/r] =>KE total = 0.75*m*vf^2 = 1857.901 J

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