Rotational inertia of nonuniform cylinder

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 9K views
physicsklutz
Messages
3
Reaction score
0
A cylinder of mass M and radius R smoothly rolls from rest along a ramp and onto a final horizontal section. From there it rolls off the ramp and lands on a floor at a horizontal distance of d = 0.505 m from the end of the ramp. The initial height of the cylinder is H = 0.98 m; the height h of the ramp is 0.10 m. The cylinder consists of an outer cylindrical shell with a certain uniform density (mass per unit volume) that is glued to a central cylinder with a different uniform density. The rotational inertia of the cylinder can be expressed in the general form I = XMR2, but is not 0.5 as for a cylinder with a single uniform density. Determine X.

I realize I should say how I attempted this problem, but that's the problem - I don't know how to attempt it. Sorry... and thanks to anyone who can help!
 
Physics news on Phys.org
Start off backwards by doing a kinematics analysis. From a ramp height of 0.10 meters, what must its horizontal velocity be in order to achieve landing on the floor 0.505 meters away. Save this initial velocity value.

Reapproach the conservational system involving the cylinder and ramp. By conservation of energy, we have:

PE1 + KE1 = PE2 + KE2

The trickey part if you'd like to say is in knowing that

[tex]KE_2 = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2[/tex]

where omega, the angular velocity = v/R.

Good luck.
 
Wait, I do not understand the first part.

"Start off backwards by doing a kinematics analysis. From a ramp height of 0.10 meters, what must its horizontal velocity be in order to achieve landing on the floor 0.505 meters away. Save this initial velocity value."

How do I go about finding the horizontal velocity?