Rotational inertia of nonuniform cylinder

In summary, the problem involves a cylinder with mass M and radius R rolling smoothly from rest along a ramp with height h = 0.10 m onto a horizontal section and landing on the floor at a distance of d = 0.505 m. The cylinder consists of an outer cylindrical shell with a uniform density and a central cylinder with a different uniform density. The rotational inertia of the cylinder can be expressed as I = XMR^2, but it is not 0.5 as for a cylinder with a single uniform density. To solve the problem, one must first determine the initial horizontal velocity needed for the cylinder to land on the floor and then approach the conservation of energy equation to find the value of X.
  • #1
physicsklutz
3
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!
 
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  • #2
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.
 
  • #3
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?
 

FAQ: Rotational inertia of nonuniform cylinder

What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution of the object and the axis of rotation.

How is rotational inertia calculated for a nonuniform cylinder?

The rotational inertia of a nonuniform cylinder can be calculated by integrating the mass distribution of the cylinder along its axis of rotation. This can be done using the formula I = ∫r²dm, where I is the rotational inertia, r is the distance from the axis of rotation, and dm is the mass element.

What factors affect the rotational inertia of a nonuniform cylinder?

The rotational inertia of a nonuniform cylinder depends on its mass distribution, axis of rotation, and the shape of the cylinder. Generally, the more mass an object has farther away from the axis of rotation, the higher its rotational inertia will be.

How does rotational inertia affect the motion of a nonuniform cylinder?

The higher the rotational inertia of a nonuniform cylinder, the harder it is to change its rotational motion. This means that objects with higher rotational inertia will require more torque to start rotating or to change their rotation speed.

How is rotational inertia of a nonuniform cylinder used in real-life applications?

The concept of rotational inertia is used in many real-life applications, such as designing vehicles, machinery, and sports equipment. It is also an important factor in understanding the stability and maneuverability of objects in motion.

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