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Erich Poppitz

PHY254 Lecture 3 ▯ Readings: Morin 3.2-3.4. Continue reading compwiki website. ▯ In this lecture we cover an example of F=ma that is not static, then consider F=ma problems in a more general format using ODEs (Ordi- nary Di▯erential Equations). Setting up F = ma problems: dynamics ▯ When the sum of the forces on a system do not balance, we get a net acceleration according to Newton’s second law: ~ dv F = m~ a = m dt (1) ▯ Like in statics, we ▯rst need to draw an FBD and decompose the forces into components. We also have to use information about constraints like string. We will not do dynamical torque problems until later in the course. ▯ Question: Consider the pulley system with masses m 1nd m (2orin, p. 59) Find the acceleration of the masses and the tension in the string. ▯ Partial Answer: The force equations are: 1 For m 2 T ▯2m g =2m a 2 2 For P 2the pulley attached to m ): 22 = T 1since2the mass of the pulley is 0). For m 1 T ▯1m g =1a . 1 Plugging in the second equation into the ▯rst equation we get the fol- lowing 2 equations: T 1 m g 1 m a ;1 1 ▯ m 1 = m 2 2 2 (2) How many equations in how many unknowns? 2 equations and 3 un- knowns. How do we close the system: we use "conservation of string", which we assume is inextensible. If m moves down 2d, m moves up 1 2 d. So a 1 ▯2a ; 2 which is the third equation we need. See the book for the rest of the solution. The ▯nal answers are: (2m 1 m )g2 2(m 2 2m )g1 2m (1 ▯ 2m )g 1 a2= ; a1= ; T = +m g1 (4m +1m ) 2 (4m 1 m ) 2 (4m 1 m ) 2 ▯ This is an example of how using constraints provides required extra information. There are lots of examples in the book. F=ma as an ODE ▯ In the previous examples, forces were usually constant which made them easy to deal with. You will notice that the accelerations were always constants. This meant it would then be easy to ▯nd the velocity and position by using constant acceleration kinematics. ▯ In general, forces can depend on variables such as time, position and velocity. We want to develop a way of solving an F=ma problem for a general F. Lets consider the one-dimensional case (or you can consider this as a single component of the vector case). In one dimension, we can write F = F(t;x;v) (3) 2 ▯ Newton’s second law states that dv m = F(t;x;v) (4) dt To get to a solution for v(t) and x(t), we are going to need to do 2 integrations, which means we will have 2 constants of integration. These can be solved for using initial conditions on the initial position and/or the initial velocity. Let’s assume the initial conditions to our above problem are given by x(t0) = x 0 v(t0) = v 0 ▯ Equation (4) has no general solution, but we can consider three rela- tively simple cases: a time dependent force, a position dependent force, and a velocity dependent force. Case 1: Time dependent force F=F(t) ▯ In this case we have: dv m = F(t) (5) dt We can separate variables mdv = F(t)dt (6) Then we can integrate the equation. Z v(t) Z t 0 0 0 m v dv = m(v(t) ▯ v ) 0 t F(t )dt ) 0 0 Z t v(t) = v + 1 F(t )dt 0 (7) 0 m t0 This gives us v = v(t). But since v(t) = dx, we can integrate again:
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