Work Energy Principle (1 Viewer)

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1. The problem statement, all variables and given/known data

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?

2. Relevant 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##


3. The attempt at a solution
 

PhanthomJay

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1. The problem statement, all variables and given/known data

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?

2. Relevant 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##


3. 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
 
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

Science Advisor
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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.
 

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