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# notes - 1 Fundamentals Exam review material ME274 ME 274...

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Fundamentals Exam review material 1 ME274 ME 274: Basic Mechanics II Dynamics Fundamentals: Mathematics, Vectors, Point Kinematics and Free Body Diagrams 1. Scalar kinematics a) Differentiation b) Integration c) Sample problems 2. Vector operations a) Addition and subtraction b) Dot (scalar) products c) Cross (vector) products d) Moments about points e) Projections of vectors f) Writing a vector in terms of a new set of unit vectors g) Sample problems 3. Vector kinematics a) Cartesian components b) Path components c) Polar components d) Sample problems 4. Free body diagrams 5. Solutions of sample problems – provided at the end of the notes

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Fundamentals Exam review material 2 ME274 Scalar kinematics: differentiation Suppose that a point P travels on a straight-line path (rectilinear motion) whose position on the path is given by the distance variable s. The speed v of P is given by: v = ds/dt . Case 1 : On this path, we are given the speed v of point P in terms of time, t: v = v(t) . From this we want to determine the acceleration of the point. The acceleration can be found by directly differentiating v with respect to t ; that is, a = dv dt Case 2 : On this path, we are given the speed v of point P in terms of its position s : v = v(s) . Here we need to employ the chain rule of differentiation (see below). Using the chain rule of differentiation, we see that: a = dv dt = dv ds ds dt = v dv ds where v = ds/dt . Chain rule of differentiation Suppose y = y(x) where x = x(t). The derivative of y with respect to t can be found from the chain rule of differentiation to be: dy dt = dy dx dx dt s v
Fundamentals Exam review material 3 ME274 Scalar kinematics: integration Again, suppose that a point P travels on a straight-line path (rectilinear motion) whose position on the path is given by the distance variable s. The speed v of P is given by: v = ds/dt . Case 1 : On this path, we are given the acceleration of point P in terms of time, t : a = a(t) . From this we want to determine the speed of the point at some instant in time t . The acceleration can be found by directly integrating a with respect to t ; that is, a = dv dt ! a t ( ) dt 0 t " = dv v 0 ( ) v " ! v t ( ) = v 0 ( ) + a t ( ) dt 0 t " Case 2 : On this path, we are given the acceleration a of point P in terms of its position s : v = v(s) . Here, prior to using separation of variables, we need to use the chain rule of differentiation to produce: a = dv dt = dv ds ds dt = v dv ds where v = ds/dt . Now, we can write: vdv = ads ! vdv v 0 ( ) v " = ads 0 s " ! 1 2 v 2 # 1 2 v 2 0 ( ) = ads 0 s " s v

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Fundamentals Exam review material 4 ME274 Scalar kinematics: sample problems Example 1 Suppose that the speed of P is given by: v t ( ) = 3 t 2 . Find the acceleration of P. Example 2 Suppose that the speed of P is given by: v s ( ) = 5sin 3 s ( ) . Find the acceleration of P. Example 3 Suppose that the acceleration of P traveling on a straight-line path is given by: a t ( ) = 6 sin 2 t ( ) , with v 0 ( ) = 4 . Find the speed of P as a function of time.
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