HeLp plz

[piper210_355]

I need some help w/ this prob:

 

A solid disk of mass 1.9kg and radius .05m rolls w/o slipping along a horizontal surface w/ a translational speed of .240 m/s. It comes to an incline that makes a 35 degree angle with the horizontal surface. Neglecting energy losses due to friction, to what height above the horizontal surface does the disk rise on the incline?

 

So the problem gives me the following variables:

mass:1.9kg

r:.05m

v:.240m/s

angle:35degrees

 

I'm not sure how to proceed though. What equation(s) should I use?

[Cap'n Refsmmat]

As soon as it reaches the incline, there will be only one force acting upon it: gravity. Break that gravitational force down into its components (the one opposing the motion and the one perpendicular to it) and you should be able to calculate how quickly gravity will decelerate the disk. Remember F = ma.

[Kyrisch]

I think it would be much easier to do this problem in terms of energy conservation. The kinetic energy of the disc [math]\frac{1}{2}I\omega^2[/math] equals the gravitational potential energy it will gain as it comes to rest at some height [math]h[/math] above the surface [math]mgh[/math]. Does that help?

[piper210_355]

okay so i'm assuming the I is inertia which is found using (1/2)mr^2

 

so

I= (1/2) (1.9) (.05)^2

I= .002375

 

 

but before i can put this in the equation you gave me i need to find out what the angular velocity is. What equation can I use for that? I looked on the internet and found one that said i could divide the angle by time but the equation doesn't give me time.

[Kyrisch]

You have the radius of the disc, which means you can determine the circumference. You also have the translational speed.

 

[math]\frac{v (m/sec)}{circumference(m)} = revolutions/sec[/math]

 

[math]1 (revolution) = 2\pi (radians)[/math]

 

With the proper conversion, you can find radians/sec which is angular velocity omega.

[Jacques]

The energy of the rotating disk is the rotational energy like Kyrisch said, plus the translational energy.

[Kyrisch]

The energy of the rotating disk is the rotational energy like Kyrisch said, plus the translational energy.

 

Ah yes, the translational kinetic energy. So, to clarify, the sum of the two kinetic energies will equal the total gravitational potential energy gained (under ideal conditions).