What Is the Linear Speed of a Disk on an Inclined Plane?

[Jbreezy]

Homework Statement

A solid disk of mass 2.5 kg with a radius of 11.0cm is released from rest on an incline plane. If the plane is 1.25 m high and the angle is 25 degrees what will be the linear speed of the disk at the bottom of the incline?
Idisk = 1/2 mr^2 and Ihoop = mr^2

Homework Equations

mgh = 1/2 mv^2 + 1/2Iw^2

The Attempt at a Solution

I = 1/2mr^2
mgh = 1/2 mv^2 + 1/2(1/2mr^2)w^2

mass cancles
gh = 1/2 v^2 + 1/4r^2w^2

substitute w = v/r

gh = 1/2 v^2 + 1/4r^2(v^2/r^2)

The radius will cancel you combine the velocity fraction get to v and you end up with something of the form

Sqrt(4/3gh) = v
Plug in the numbers and you get 4.04 m/s

Is this right? Please help me double check. I felt there was a lot of information that did not get used but that doesn't mean it is not correct.
Thanks,
j

 

[Doc Al]

Looks good to me.

 

[Jbreezy]

OK, well the more eyes the better. Thanks dude. It is just I feel he gave information to throw me off or something. I don't know.

 

[Doc Al]

It is just I feel he gave information to throw me off or something. I don't know.

Part of knowing how to solve a problem is knowing what's important and what's not.

And there are several ways to solve for the speed. What you did, solving it symbolically and only plugging numbers in at the last step, is the smart way. Someone else might have plugged numbers in at every step. The "unused" data would have been used by such a person.

 

[FermionXLR8r]

Yeah, It is right.

 

[Jbreezy]

Hi I have another one posted with torque and hanging a sign if someone oone could double check me it would be great. thanks for the help.