Why Does a Rolling Disk Have No Translational Kinetic Energy?

[princejan7]

Homework Statement

http://postimg.org/image/m9wtlg5ah/

taking T1 for example
why does the disk have no translational kinetic energy component?

And why does the friction do no work?

Homework Equations

Kinetic energy of a rigid body in planar motion

T = $\frac{1}{2}m{v_G}^2 + \frac{1}{2}I_G{w_B}^2$

The Attempt at a Solution

 

[PhanthomJay]

Homework Statement

http://postimg.org/image/m9wtlg5ah/

taking T1 for example
why does the disk have no translational kinetic energy component?

And why does the friction do no work?

Homework Equations

Kinetic energy of a rigid body in planar motion

T = $\frac{1}{2}m{v_G}^2 + \frac{1}{2}I_G{w_B}^2$

The Attempt at a Solution

Since the disc is not slipping, the friction force is static, not kinetic. Being static , the contact is always instantaneously at rest with respect to the ground, and since it doesn't move, the friction does no work.
Regarding translational KE, well, it exists for sure, but the solution chose to calculate the moment of inertia of the disc about the ground, not its mass center, and by so doing, it already encompasses the translational component of the KE

 

[princejan7]

Since the disc is not slipping, the friction force is static, not kinetic. Being static , the contact is always instantaneously at rest with respect to the ground, and since it doesn't move, the friction does no work.
Regarding translational KE, well, it exists for sure, but the solution chose to calculate the moment of inertia of the disc about the ground, not its mass center, and by so doing, it already encompasses the translational component of the KE

thanks
could you explain why using $I_q$ encompasses the translational component of the KE

 

[PhanthomJay]

thanks
could you explain why using $I_q$ encompasses the translational component of the KE

The translational kinetic energy is encompassed in the 'mr^2' term of the parallel axis theorem. When using the contact point as the axis of rotation, recall that the instantaneous velocity of that point with respect to the surface is 0 , so in effect the object is considered rotating about that point in pure rotation. It is an alternate means of approaching the problem. Consider for example a rolling disk of mass m and radius r, rolling on a level surface without slipping. You can calculate its kinetic energy by summing the rotational KE through the center of mass and the translation KE of its center of mass, or you can use the parallel axis theorem to determine the rotational inertia about the contact point and calculate the KE as pure rotational KE without the translational KE (which is 0). You will get the same result either way.

 

[HalfLostAndSearching]

The Work Energy Principle states that the work done on a system is equal to the change in kinetic energy of the system. In this case, the disk is initially at rest and therefore has no initial translational kinetic energy. As the disk rolls down the incline, it gains rotational kinetic energy due to its angular velocity, but it does not gain any translational kinetic energy because it is constrained to roll without slipping. This is due to the fact that the point of contact between the disk and the incline is always at rest, and thus there is no relative motion between the two surfaces. Therefore, the disk does not have a translational kinetic energy component.

As for the friction, it does not do any work because it acts perpendicular to the direction of motion of the disk. Work is only done when a force acts in the same direction as the displacement of an object. In this case, the force of friction acts in the opposite direction of the displacement of the disk, and therefore does not do any work. This is consistent with the Work Energy Principle, as the change in kinetic energy of the system is solely due to the gravitational force acting on the disk.