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Discuss the motion of a disc rolling without slipping on a level surface. Hence find the condition for rolling without slipping? |
Answer» Solution :Let us consider a disc that rolls without slipping on a level SURFACE. Let the radius of the ROLLING object be R in one full rotation. The centre of mass is displaced by `2piR`. We would agree that not onoy the centre of mass, but all the points on the disc are displaced by the same `2piR` after one full rotation. The only difference is that the centre of mass takes as straight path, where as, all the other points undergo a path which has COMBINATION of the translational and rotational motion. Especially the point on the edge undergoes a path of a cycloid as shown in the diagram.  As the centre of mass takes only a straigh line path, its velocity `v_(CM)` is only translation velocity `v_("TRANS")(V_(CM)=V_("TRANS"))`.All the other point have two velocities. One is the translational velocity `v_("TRANS")` and the other is rotational velocity `v_("ROT")(V_("ROT")=romega)`. Here r is the distance of the point from the centre of mas and `omega` is the angular velocity. The rotational velocity `v_("ROT")` is perpendicular to the instantaneous position VECTOR from the centre of mass as shown in the figure (a). The resultant of these two velocities is v. This resultant v is perpendicular to the position vector from the point fo contact of the rolling object with the surface on which it is rolling as shown in the figure.  Importance of point of contact: In pure rolling the point of the rolling object that comes in contact with the surface is at momentary rest. This is the case with every point that is one the edge of the rolling object. As the rolling PROCEEDS, all theponts on the edge, one by one come in contact with the surface. the points remain at momentary rest at teh time of contact and then take the path of the cycloid as already mentioned. Hence Let us consider the pure rolling in two different ways. (i) The combination of translational motion and rotational motion about the centre of mass (or) (ii) The momentary rotational motion about the point of contact. As teh point of contact is at momentary rest in pure rolling, its resultant velocity v is zero `(v=0)`. For example in the figure, at the point of contact `v_("TRANS")` is forward ( to right) and `v_("ROT")` is backwards.  Which implis that `v_("TRANS")` and `v_("ROT")` are equal in magnitude and opposite in direction `(v=v_("TRANS")-v_("ROT")=0)`. Hence let us conclude that in poure rolling, for al the points on the edge, the magnitudeof `v_("TRANS")` and `v_("ROT")` are equal `(v_("TRANS")=v_("ROT"))`. As `v_("TRANS")=v_(CM)` and `v_("ROT")=R omega`, in pure rolling we have `v_CM)=R omega` ...........1 The special feature of the equation are 1. in rotational motion, as per the relation `v=r omega`, the centre point will not have any velocity as r is zero. but in rolling motion, it suggests that the centre point has a velocity `v_(CM)` given by equation 1.  At the topmost point, the two velocities, `v_("TRANS")` and `v_("ROT"0` are equal in magnitude and in the same direction (to the right). Thus the resultant velocity v is the sum of these two velocities, `v-v_("TRANS")+v_("ROT")` . In other form `v-2v_(CM)` as shown in the figure. |
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