2
$\begingroup$

So this is the problem:A wheel of radius of gyration k is placed on a belt moving with a speed v, which is maintained constant by means of an external agency. Assume that the axis of the wheel is fixed, so that the wheel is free to rotate about its axis perpendicular to its plane through the centre. We are required to find the energy supplied by the external agency till slipping ceases.

Now this might be due to two reasons, due to increase in kinetic energy of the wheel, and also the energy expended to move the belt at constant speed. My question is, are these two really the same, so that I need to count only one of these, or are these independent of each other. What I think is that the friction does some negative work on the system, and the exact negative of this work is the energy expended. So I need to sum the positive work on the wheel and the negative work on the belt, and the negative of this work will be the energy expended.Is this reasoning correct?

$\endgroup$
2
$\begingroup$

When the wheel comes in contact with the belt friction will act as there will be relative motion between the belt and the point of contact. Now friction will tend to act on the belt opposite to the velocity of the belt until slipping ceases.

To find the work done by the external agency lets consider the energy changes : 1) The K.E of the wheel increases. 2) As sliding is there some energy will be dissipiated as heat.

Taking the belt and the wheel as our system :- The energy supplied by the external agency = K.E. of Wheel + Heat generated.

Also since we only need to move the belt at constant velocity the work done by friction on the belt must be equal (in magnitude and opposite in sign) to the work done by the external agency on the belt as there is no change in the kinetic energy of the belt .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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