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Thursday, November 1, 2007

Question for Hw 6/79

Is the answer in the book for 6/79 correct? I followed the sample problem in the book (6/5) that is similar and obtained a=g/2*sin(theta), but got a different answer in than in the book, am I missing something?

8 comments:

Lenka Kollar said...

Nevermind, I figured out what I did wrong.

Andrew Crandall said...

Mass Moment of Inertia for a cylinder rolling about it's z-axis...1/2mr^2? That's what I'm using but I'm not getting the answer. The only force I have creating a nonzero moment about G is friction, and from my kinematics analysis I have a_G = alpha*r [in the 'i' direction, my x axis is aligned with the incline). I don't seeing where I'm going wrong.

Jerad said...

check your Ig (mass moment of inertia) value.

KMP said...

Is the Mass Moment of Inertia 1/2mr^2 or just mr^2?

Andrew Coury said...

The mass moment of inertia for a solid cylinder is 1/2mr^2. I think the problem you are having is that you are trying to calculate the moment about G using F = (mu)*N. But remember that this is only the maximum frictional force. If the cylinder rolls without slipping you could have a frictional force less than this. I suggest instead you take the moment about the contact point, C, and use [Sigma]Mc = I[alpha] + mad. This will avoid the friction force altogether.

Matt.K said...

Mad? what is point d in Your FBD?

Andrew Coury said...

What I meant was m*a*d, where d is the perpendicular distance between the force and the point you are taking the moment about. In this case d = r = 6in.

Andrew Coury said...

The equation in the book is 6/2.