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?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 :)
how about now
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)
Not the answer you are looking for? Search for more explanations.
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
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
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)