What is the cylinder's angular velocity when it is directly below the axle?

[crudities360]

A 5.0 kg, 60-cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.

 

What is the cylinder's angular velocity when it is directly below the axle?

[Kosar_For_President]

A 5.0 kg, 60-cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.

 

What is the cylinder's angular velocity when it is directly below the axle?

 

2

[Mr. T]

it depends on several variables

[calfoxwc]

You're supposed to do you own homework...

[VaporTrail]

Oh shit. It's been awhile since I took mechanics. Okay, you know that it's speed is going to be dependent on gravity. So F in the j-hat direction is = -ma

 

a = 9.8 m/s^2

 

Okay, I think you can simplify the problem by treating the disk as a point mass. The only radius that is important is that of the center of mass. So if it's a uniform disk, the center of mass is in the middle, which is 30cm from the point of rotation.

 

 

._________O

 

^^^^So basically in the line above, the period is the point of rotation, the line is a massless, stiff wire, and the O is your mass.

 

you need to know the following

 

x=r cos (theta)

y=r sin (theta)

 

Where r is the radius of the wire, theta is the angle between level and the wire.

 

You'll have to use one of those with the angular acceleration

 

And then.

 

Use

 

 

You're going to need to know v initial. I'm not gonna help you any more unless you show me what work you've done.

 

DISREGARD THAT, I SUCK COCKS

 

lol, god I'm terrible at mechanics. least interesting subject in physics, imo.

 

Are you a physics major?

 

http://www.physicsforums.com/showthread.php?t=168974