Moment of Inertia 2: Kinetic Energy, Speed & Acceleration

In summary, a rolling disk with radius R and mass M on a level surface has a kinetic energy of T = .5mv^2 + .5Iw^2, where I is the moment of inertia about an axis perpendicular to the page and through the center of the disk. The instantaneous speed vL, relative to the ground, of the lowest point on the disk is equal to v/R, while the instantaneous speed vH, relative to the ground, of the highest point on the disk is equal to -v/R. When the disk is allowed to roll down an inclined plane at an angle θ relative to the horizontal, the magnitude of its center of mass acceleration is a = 2/3 gsinθ, assuming
  • #1
joemama69
399
0

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)

Wheight = v/R
Wbottom = -v/R

part c)

mgsinQ - f = ma

fr = .5mr2[tex]\alpha[/tex]

mgsinQ - .5ma = ma

gsinQ = 3/2 a

a = 2/3 gsinQ
 

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  • #2
joemama69 said:
part a)

T = .5mv2 + .5Iw2

is this right, its barley even a question

Yup, that's right.

part b)

Wheight = v/R
Wbottom = -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[tex]\alpha[/tex]

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
 
  • #3



I would like to confirm that the equations used in the attempt at a solution for parts a and b are correct. The equation for kinetic energy (T) is indeed correct, as it accounts for both the translational and rotational kinetic energy of the disk. Additionally, the equations for the instantaneous speeds at the lowest and highest points of the disk are also correct, as they take into account the radius of the disk (R) and its rotational speed (w).

For part c, the equation used to calculate the acceleration of the center of mass is also correct, as it takes into account the forces acting on the disk (gravity and friction) and the mass of the disk. However, it is important to note that the angle (theta) in the equation should be the angle of the inclined plane, not the angle relative to the horizontal. Additionally, the frictional force (f) should be equal to the coefficient of friction (mu) multiplied by the normal force (mgcos(theta)), rather than just being equal to the normal force as shown in the equation. Therefore, the correct equation would be:

mgsin(theta) - mu*mgcos(theta) = ma

Solving for acceleration (a) would give:

a = g(sin(theta) - mu*cos(theta))

Overall, the attempt at a solution shows a good understanding of the concepts of moment of inertia, kinetic energy, speed, and acceleration. However, it is important to pay attention to the details and use the correct equations for each scenario.
 

Related to Moment of Inertia 2: Kinetic Energy, Speed & Acceleration

What is Moment of Inertia?

Moment of Inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution and the shape of the object.

How is Moment of Inertia calculated?

The formula for Moment of Inertia is I = m * r^2, where I is the moment of inertia, m is the mass of the object, and r is the distance from the axis of rotation to the object.

How does Moment of Inertia affect an object's kinetic energy?

The Moment of Inertia affects an object's kinetic energy by determining how much rotational kinetic energy the object will have for a given angular velocity. The higher the Moment of Inertia, the more energy is needed to accelerate the object.

What is the relationship between Moment of Inertia and speed?

The relationship between Moment of Inertia and speed is inverse. This means that as the Moment of Inertia increases, the speed of the object decreases, and vice versa.

How does Moment of Inertia impact an object's acceleration?

The Moment of Inertia affects an object's acceleration by determining the torque required to produce a given angular acceleration. The higher the Moment of Inertia, the more torque is needed to accelerate the object.

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