Question

Name Hakees Mrehealy Ha Hee Table: Section: 2 Discussion: Rotational Dynamics A block of mass 2m hangs from a string wrapped around a pulley of mass m The pulley is a uniform solid disk of radius R rotating without friction about the center, so other helpful relationships:… and K- The system is released from rest. First, we'll apply energy ideas. 2m Approach 1: define the system as the Earth, the block, the pulley, and the string ystem-inicial energy ystem fal eer The energy bar graphs are shown for the initial situation. Sketch the energy bar graphs for the final situation, after the system has been released from rest and the block has dropped through a height h, where the final potential energy is zero. Hint: Write out the five-term energy equation for the system. b block p puley Approach 2: Now, we'll look at two separate pieces of the system, just Block-final enery E Puley final energ the block by itselfand just the pulley by itself. Again, sketch energy bar graphs for these two objects after the system has been released from rest and the block has dropped through a height h. Note that the initial energy bars would all be zero. Based on the bar graphs, if the force of gravity acting on the block is 2mg, how large is the force of tension, in terms of mg? w Ws work done by the force of gravity W,-work done by the force of tension Boston University Studio Physics-Rotational Dynamics Discussion page 1

Answer 1

Mass of pulley = m mass of block = 2m Approach 1 U_f = 0 U_i + k_i + W_nc = U_f + k_f mgh + 0 + 0 = 0 + 1/2 2 mv^2 + 1/2 mR^2/2 v^2/R^2 mgh = 5/4 mv^2 equation (1) Final kinetic energy of pulley k_p = 1/2 Iw^2 = mR^2/4 v^2/R^2 = mv^2/4 = mgh/5 Final kinetic energy of block k_b = 1/2 times 2mv^2 = 4/5 mgh Approach 2 on Block W_g = mgh final K.E = K_b_f = 4/5 mgh k_b_i = 0 W_g + W_t = K_b_f - K_b_i mgh + W_t = 4/5 mgh - 0 W_t = -mgh/5 on pulley W_t = K_p_f - k_p_i W_t = mgh/5 - 0 W_t = F_t h = mgh/5 F_t = mg/5 rightarrow Tension force