Question

Suppose the height of the ramp is h1= 0.3 m, and the foot of the ramp is horizontal, and is h2= 1.00 m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 12.7 mm in each case; assume that the density of steel is 7.8 g/cm^3; and assume that the density of aluminum is 2.7 g/cm^3.

a) A solid steel sphere sliding down the ramp without friction
b) A solid steel sphere rolling down the ramp without slipping.
c) A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping
d) A solid aluminum sphere rolling down the ramp without slipping.

There some formula given which summarize all the way down to
-2sqrt(h1h2) - sliding object
-1.690sqrt(h1h2) - rolling solid sphere
-1.549sqrt(h1h2) - rolling hollow sphere

but i don't think were suppose to use them since i don't get why we don't use the density of thickness of the sphere

Answer 1

Time to fall for all objects is t = sqrt(1/4.9)
Now find the speeds of the different objects at the bottom.
For a) it's a simple mgh = 1/2mv^2 (h is the ramp height in this case)
For the other 3 cases mgh = 1/2mv^2 + 1/2I(omega)^2 The I's you'll have to look up, they're related to mass and radius of the objects (so you do need density). The omega is just v/r.
Hope that helps.