physics7a-fall03-final-Yu-soln

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question 2 (10 Points) Energy & Work Figure 2 shows a massive spring whose total mass is M s and whose length when unstretched is L. A smaller mass m is attached to the end of the spring. The mass m and the spring both move on a frictionless table. The spring, when stretched, behaves like a rod with Young’s modulus E. When the end of the spring is stretched by an amount D the extension of a small element of the spring around the point x can be assumed to be given by (x/L)D. Similarly when the mass m at the end of the spring is moving with velocity V a point x along the spring can be assumed to move with velocity (x/L)V. Fig.2 (a) Calculate the potential energy of the mass m plus the spring when m is displaced by D. (b) Calculate the kinetic energy of the mass m plus the spring when m is moving with V. (c) What is the frequency of oscillation of m? Solution (a) For a rod of length L, area of cross section A and Young’s modulus E the stress (=Force F/Area A)=E(strain) where strain is defined as L/L. Consider the point x along the spring. When it is extended by amount x the restoring force on it is: F=EA( x/x). x is given by: x = (x/L)D so : F=EA(D/L). The direction of F is in the negative x direction. Notice that this force F is independent of x just like the tension in a string is the same everywhere. Thus the spring constant k of the massive spring is =EA/L. The work done by external force in stretching the end of the spring by the amount D is then given by the integral of Fd l (where d l is the displacement of the end of the spring) from l =0 to D, just like a regular massless spring. Thus the potential energy stored in the spring is : (1/2)kD
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online