Tuesday, December 16, 2008

Irodov Problem 1.260

Since the cylinder if affixed to the mount, due to the application of the force F, both of them will move together and can be treated as one body. If the acceleration of the mount+cylinder system is a then based on Newton's laws,

The force F also induces a torque of Fr on the cylinder since it acts at a distance r from the axis of rotation. Since the moment of intertia of the cylider about its axis of rotation is , we have,

This rotation of the cylinder unwraps the wound thread on the cylinder and releases it, thus point K moves away from the mount+cylinder system with an acceleration of . The net acceleration of point K as seen by a stationary observer is the sum of i) the acceleration of the mount+cylinder system a, and ii) the acceleration of point K as seen by an observer on the mount . Thus, from (1) and (2) we have the acceleration of point K as,


arhar jain said...

sir, i have a doubt in prob. 1.260

i think there is one more force which has not been taken into account. The hinge reaction between the mount and the cylinder is responsible for bringing accelaration to the mount. but this force has not been taken into account while writing the eqns. plz help know if i'm wrong or right. it would help clearing my confusion.

Krishna Kant Chintalapudi said...

Dear Arhar,

when you consider, the cylinder and the mount as a whole body, then you need not consider internal reaction forces since they cancel out. At the hinge there are in fact two forces, the mount exerts force on the cylinder and the cylinder exerts force on the mount and they are equal and opposite. So when you consider the entire system as a whole these two forces cancel each other.

If you draw the free body diagram of the mount only, however, you will see that the hinge exerts a force on the mount. Similarly if you draw a free body diagram of the cylinder , the hinge will be exerting a force on the cylinder in the opposite direction.

Hope this helps.