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sirenitabaila
a solid sphere is started from rest at the top of an almost frictionless incline..with just enough friction to keep it rolling but not enough to include in an energy conservation calculation. I=2/5(MR^2), what is the sphere's speed when it has rolled to the end of the incline? height from ground = 1m, end height = .05 from ground So far, I have rolling object, K1 = 1/2(MV^2) + 1/2(IW^2) use energy conservation? K1 = K2?
im thinking rotational kinematics equations to find displacement/final velocity? mostly just trying to think of this conceptually rather than specifics that i can work out on my own and need the practice :)
Yes you can solve this by using energy conservation since you're given the fact that there is just enough friction to keep it rolling at constant speed but not enough to be included in energy calculation. energy at start when there is no velocity (since we're starting from rest) all v terms = 0 So energy at start = mgh (potential energy) Energy at the end = 1/2mv^2 (kinetic) + 1/2 mr^2(w)^2 (inertial energy) if we put these equal to each other, which is the conservation of energy law mgh = 1/2m(v)^2 + 1/2mr^2(w)^2 gh = 1/2v^2 + 1/2r^2(w^2) (all mass terms cancel out) Also, w (angular velocity) = velocity/radius, w = v/r, so w^2 = V^2/r^2 gh = 1/2v^2 + 1/2r^2(V^2/r^2) gh = v^2 v = root (gh)
loook at this elegant solution
the energy mgh gets transferred to rotational and translational kinetic energy mgh = 1/2 mv^2 + 1/2 Iw^2 putting i =2/5mr^2 we get mgh = 1/2mv^2 + 2/10 mv^2 so u can clearly see that 3/5 th of the mgh goes to translational and 2/5 th of the mgh goes to rotational 1/2mv^2 = 3/5 mgh v ={6/5gh}^1/2
oops forgot the 2/5 for Iw^2
I made a mistake the translational and rotational comonents gets divided in the ratio 2 ist 5, so using basic ratio and proportions we get translational component = 5/7 mgh hence v= root(5/7 gh)