USAPhO 2014 F=ma exam #14, (rotational dynamics)

[Agrasin]

PDF: http://www.aapt.org/physicsteam/2014/upload/exam1-2014-2-2-answers.pdf

Homework Statement

A disk of moment of inertia I, mass M, and radius R has a cord wrapped around it tightly as
shown in the diagram. The disk is free to slide on its side as shown in the top down view. A
constant force of T is applied to the end of the cord and accelerates the disk along a frictionless
surface.

After the disk has accelerated some distance, determine the ratio of the translational KE to total KE of the disk,

KE_{translational} / KE_total =

Answer) I / (MR² +I)

Homework Equations

Torque = Iα = F x R

KE = 1/2 mv² or 1/2 Iω²

The Attempt at a Solution

My conceptual understanding is the problem here, I think. If there is absolutely no friction, then would there be any rotation? I think there would be, but I'm not sure.

Here's an attempt:

KE_{translational} / KE_total = Mv² / (Mv² + Iω²)

Substituting v = Rω, cancelling the ω² terms

= MR² / (MR² + I)

So I get an answer close but not exactly the correct answer. The answer I got is choice E on the actual exam, meaning my attempt probably has a common mistake.

 

[Dick]

PDF: http://www.aapt.org/physicsteam/2014/upload/exam1-2014-2-2-answers.pdf

Homework Statement

A disk of moment of inertia I, mass M, and radius R has a cord wrapped around it tightly as
shown in the diagram. The disk is free to slide on its side as shown in the top down view. A
constant force of T is applied to the end of the cord and accelerates the disk along a frictionless
surface.

After the disk has accelerated some distance, determine the ratio of the translational KE to total KE of the disk,

KE_{translational} / KE_total =

Answer) I / (MR² +I)

Homework Equations

Torque = Iα = F x R

KE = 1/2 mv² or 1/2 Iω²

The Attempt at a Solution

My conceptual understanding is the problem here, I think. If there is absolutely no friction, then would there be any rotation? I think there would be, but I'm not sure.

Here's an attempt:

KE_{translational} / KE_total = Mv² / (Mv² + Iω²)

Substituting v = Rω, cancelling the ω² terms

= MR² / (MR² + I)

So I get an answer close but not exactly the correct answer. The answer I got is choice E on the actual exam, meaning my attempt probably has a common mistake.

v = Rω isn't true. That's only true if you are rolling without slipping. Since you have no friction, that's not true. Use F=ma=mdv/dt and a similar form for the rotational motion to compute both forms of energy as a function of time, t.