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physics7a-fall03-final-Yu-soln

physics7a-fall03-final-Yu-soln - Question 1(10 Points 1D...

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Question 1 (10 Points) 1D Collision and Impulse A uniform rope of mass m per unit length hangs vertically from a hook so that the lower end of the rope just touches the horizontal table as shown in Fig. 1(a). The rope is then released from the hook and descends onto the table. The rope is assumed to be completely flexible and when it reaches the table it stops instantaneously. When a length y of the rope has fallen (see Fig. 1 (b)) what is the force exerted by the rope on the table in terms of y, m and g? Fig. 1 Solution The rope descends via free fall onto the table. As it starts from zero initial velocity its velocity v is related to the height y it has fallen by: v(y)=[2gy] 1/2 . The length of the rope which lands on the table during the time dt at that instant is: v(y)dt and the corresponding mass is mvdt Based on the assumption that the rope stops instantaneously after it lands on the table, the rate at which the rope transfers momentum to the table is: dp/dt=[mvdt]v/dt=mv 2 =m(2gy). According to Newton’s equation of motion this is the force exerted on the table by the falling rope. However, there is also rope of length my already on the table and the force they exert due to gravity on the table is myg. Thus the total force exerted by the rope on the table is 2mgy+mgy=3mgy.
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