A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 4.4 m/s when it reaches a 19° incline

Question

A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 4.4 m/s when it reaches a 19° incline.

(a) How far up the incline will it go?

(b) How long will it be on the incline before it arrives back at the bottom?

Answer 1

change in KE= gain in PE, there's also friction acting on it but its total work would cancel out which is rotational+translational both in opposite directions.

Let it goes h distance on incline then PE= mgh sin 19

KE= 1/2 mv^2( 1+ 1) (I've condensed the rotational and translational KE)

since its rolling the point of contact is at rest always so you'll have v= wr where w is angular velocity and since I= mr^2 for hollow cylinder the radius cancels out in the energy equation leaving you with the above equation.

Solve to get h

For time you need to find the retardation during the incline motion.

Start by equating the forces in translational

f-mgsin 19=- ma

(f is the friction force acting upwards , that's bcoz the gravitational force will have impact on translational velocity and thus the point of contact will have a greater velocity in backward, to counter this friction acts upwards)

Now use the rotational force equation, take torque about center of the hoop then

f r= I alpha ( the relation between alpha and a is that point of contact is at rest so a=alpha r)

solve these two to get a (translational retardation)

Now since you've the distance h and the acceleration and the initial velocity v=4.4 put those in the equation x=ut+1/2 a t^2 to get t your time would be twice of that!