Gluesticks

Extremeley difficult physics question I couldn't solve for extra credit?

A long thin rod (mass M, length L) has a very small block of mass m glued to its right end. The left end is attached to a wall with a pin joint, which acts a pivot around which the rod can rotate. The rod is released from rest from a horizontal configuration. It swings down and crashes into the wall. What is its angular speed immediately before crashing ? the answer is : (g(6 m + 3M)/L(M + 3m)) ^1/2 can u explain how to get the answer ? i had this problem 3 months ago in physics hon. and only a few people knew how to do it this was in high school ....so if your good at college physics or whatever help out

Public Comments

1. hmm... so essentially what you want to use here is conservation of energy. It can get confusing though. Consider the rod in its horizontal configuration: ════════(m) <------L------> Then in its vertical configuration: ║ ^ ║ | ║ L ║ | ║ v (m) Since the rod's centre of mass is at its centre, it has dropped a distance of L/2, an energy change of gML/2. The mass m has dropped a distance L, an energy change of gmL. Thus, an energy transform of gL(2m+M)/2 occurs as GPE is converted into kinetic energy. The kinetic energy is entirely rotational in this case, since there is a pivot point. Thus, we need the moment of inertia. The moment of inertia for a thin rod (mass M, length L) about one of its ends is given by ML²/3; the point mass m at distance L from the centre of rotation trivially contributes a further mL² to the moment. Thus, the moment of rotational inertia here is: I = L²(3m+M)/3. Now the equation E_k = ½ Iω² may be usefuly applied: ½ (L²(3m+M)/3) ω² = gL(2m+M)/2 (L²(3m+M)/3) ω² = gL(2m+M) ω² = 3gL(2m+M)/(L²(3m+M)) ω² = g(6m+3M)/(L(3m+M)) ω = √(g(6m+3M)/(L(3m+M))). Q. E. D.