let h be the height above the bottom of the loop, so the excess Penergy is mg(h=2r). That energy is converted to KE, both translational and rotational.
rotational energy=2/3 mr^2*w^2
were w=v/r
rotational energy=2/3 m v^2
translational energy= 1/2 m v^2
total energy=mg(h-2r)
7/6 m v^2=mg(h-2r)
v^2=6/7 g (h-.35)
to get that velocity, find the velocity at the top.
now, to stay on the loop,
gravity=mv^2/r
mg=mv^2/r
v^2=g*r
So finally, in
v^2=6/7 g (h-.35) solve for h
now the total energy at the bottom is 1/2 mv^2(top)+mg(2r)=1/2 mvb^2
solve for vbottom.
A solid sphere and a thin walled spherical she'll, both with a radius R, are set up to roll down a ramp and through a vertical loop of radius r. From what minimum height, h, does each sphere need to be released in order to make it around the loop?
If the loop has a radius of r= 35cm what speed will each sphere have after it has excited the loop
1 answer