Solution of Tutorial Sheet 6

Solution of Tutorial Sheet 6 - Solution of Tutorial Sheet 6...

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Unformatted text preview: Solution of Tutorial Sheet 6 1. We can start from P = Fv. Hence, This leads to an ordinary differential equation (ODE) in the form of: This We can do separation of variables Since the initial condition is v=0 at t=0, we can integrate previous equation This leads to an expression of v as a function of time. 2 2 √ To get x(t) we need to integrate velocity To 22 3 To get a(t) we differentiate v(t) 12 2 If m=1000 kg, P=100 kW, and v =100 km/h=27.7 m/s, we can calculate the time needed from v(t) expression. 2 3.84 . 2. A) The relation of angle and height of mass is: cos The potential energy is: cos Therefore, the change in potential energy is: Δ cos cos 1 B) Change in potential energy is equal to the kinetic energy gain. Δ Δ 0Δ 1 cos C) From B, we get 1 1 cos 2 2 1 cos Hence, the radial acceleration is 2 1 cos For tangential acceleration, we can get it from Newton’s 2nd law easily. sin D) Utilizing Newton’s Law again Mass flies off means the normal force is zero. Hence, we get cos And since the mechanical energy is conserved 1 0 cos 2 2 cos 3 Approximately the angle is 48 degrees. 3. A) . Frictional force is opposite of direction of motion. B) . Spring pulls mass in the opposite direction of direction of motion. C) Yes, normal force and gravitational force. They do no work since their direction is perpendicular to direction of motion. cos D) Total work = E) We can treat the spring and the block as one system. There is one nonconservative force conservative doing work, that is the frictional force. At first there is only kinetic energy due to initial velocity . At the final condition some of the initial kinetic energy has been transformed into the spring potential energy (the spring is nitial stretched) and the rest of the initial energy has been lost (used up) due to work by frictional force. Hence, 1 1 2 2 1 1 0 2 2 2 0 It’s a quadratic equation, thus: 2 4 2 4 Since negative solution doesn’t have physical reason, we only use one solution (the positive ason, length) We know that the spring do work too, but why we do not put another term of as a representation of the work done by spring? The answer is because the spring potential energy representation has represent the work done, so there is no need to double calculate them. Another approach on this kind of problem is given in the textbook Chapter 8 part 3, and Example 8.6. on A summary of these two approaches will be given in a separate sheet. approach 4. In the system, mechanical energy is conserved. 1 1 0 sin 2 2 2 1.24 sin / ...
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This note was uploaded on 12/08/2010 for the course SPMS pap111 taught by Professor Claus-dieterohl during the Fall '10 term at Nanyang Technological University.

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