Rolling disk: kinetic energy, point speeds, and acceleration down an incline

[joemama69]

Homework Statement

a) A disk with radius R and mass M is rolling without slipping on a level surface, as shown. The moment of inertia about an axis perpendicular to the page and through the center of the disk is I. If the center of
the disk is moving at speed v, what is the kinetic energy of the disk?
b) For the disk described in part (a), what is the instantaneous speed vL, relative to the ground, of the lowest point on the disk? What is the instantaneous speed vH, relative to the ground, of the highest point on
the disk?

c) Suppose the disk described in part (b) is allowed to roll down an inclined plane, oriented at an angle θ relative to the horizontal, as shown.
Assume that it rolls without slipping, but ignore all frictional effects other than the friction that is necessary to prevent it from slipping. Let g
denote the acceleration of gravity. Calculate the magnitude aCM of the acceleration of the center of mass.

Homework Equations

The Attempt at a Solution

part a)

T = .5mv² + .5Iw²

is this right, its barley even a question

part b)

W_height = v/R
W_bottom = -v/R

part c)

mgsinQ - f = ma

fr = .5mr²$$\alpha$$

mgsinQ - .5ma = ma

gsinQ = 3/2 a

a = 2/3 gsinQ

 

[ideasrule]

part a)

T = .5mv² + .5Iw²

is this right, its barley even a question

Yup, that's right.

part b)

W_height = v/R
W_bottom = -v/R

The question asks for velocity relative to the ground, not for angular speed relative to the center of mass. If the bottom of the wheel is moving at v relative to the center of mass and the center of mass is moving at v relative to the ground, how fast is the bottom moving with respect to the ground? How about the top?

part c)

mgsinQ - f = ma

fr = .5mr²$$\alpha$$

mgsinQ - .5ma = ma

gsinQ = 3/2 a

a = 2/3 gsinQ

Yeah, that's right. An easier way to do things is to consider the contact point as the "pivot" and apply the rotational equations to that. "I" would be 0.5mR^2 + mR^2 = 1.5mR^2 and torque would be mgRsinQ, so:

mgRsinQ=1.5mR^2*alpha
gsinQ=1.5R*a/R
a=2/3gsinQ