0
$\begingroup$

enter image description here

Two identical discs of mass M and radius R are connected by a light rod. The assembly rests at a corner, the vertical wall is smooth and there is sufficient friction on the floor to ensure pure rolling of disc B. The system starts from position θ=0.

This is our scenario and I've been told energy is conserved in this scenario and I don't understand why that is.

Aren't the normal forces from the wall and the floor external forces ?

Link to the solution

$\endgroup$
4
  • 1
    $\begingroup$ Looking at the answer in the link: Have $V_A \cos \theta = V_B \sin \theta$. Surely the next line $\implies V_A = V_B \sin \theta$ is incorrect? $\endgroup$
    – jim
    Jan 4, 2021 at 10:45
  • $\begingroup$ Energy is always conserved so you mean "how" is energy conserved. $\endgroup$
    – my2cts
    Jan 4, 2021 at 11:30
  • $\begingroup$ @my2cts Seems fine to me sir/ma'am. Why is energy conserved here in the presence of external forces ? Is it grammatically incorrect? $\endgroup$ Jan 4, 2021 at 15:44
  • $\begingroup$ @Glowingbluejuicebox my2cts is refering to first law of thermodynamics-"energy can neither be created nor be destroyed" and indeed has a point.You must specify a system for which you are try ing to conserve energy .Energy is always conserved irrespective of your ability to represent it mathematically. $\endgroup$
    – Protein
    Jan 5, 2021 at 10:00

1 Answer 1

3
$\begingroup$

Aren't the normal forces from the wall and the floor external forces

Yes they are but the work done by the mentioned forces is zero. The important concept here is work done by friction when object is rolling .

Look at normal forces acting at both points of contact and the direction of motion of respective disks. What do you observe? They are perpendicular! Thus the work done is zero.

look here for work by frictional force in rolling motion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.