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## Final Exam Review

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**Please Return Loan Clickers to the MEG office after**Class!Today!**Review**Always work from first Principles!**Review**Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints**Review**1. Free-Body**B_x**mg B_y 1. Free-Body**B_x**mg B_y 2. Newton Moments about B: -mg*L/2 = IB*a with IB = m*L2/3**B_x**mg B_y 3. Constraint aG = a*L/2 = -g*3/4**mg**A_x A_y N 1. Free-Body**mg**2. Newton A_x A_y N Moments about Center of Cylinder:A_x From triangle at left: Ax*(R-h) –b*mg = 0 acart*(R-h) –b*g = 0**mg**2. Newton A_x A_y N N = 0 at impending rolling, thus Ay = mg Ax = m*acart**Kinematics (P. 16-126)**4r -2r*i + 2r*j CTR**Feedback**• Overall, when comparing traditional Homework formats with Mastering, I prefer • Paper submission of Homework • Electronic Submission**Feedback**• For me, the most useful benefit of Mastering is • Hints while developing the solution to a problem • Instant grading of results • Practice Exams**Point Mass Dynamics**X-Y Coordinates**A ball is thrown horizontally from A and passes through**B(d=20,h = -20) meters. • The travel time t to Point B is • t = 4 s • t = 1 s • t = 0.5 s • (D)t = 2 s • Use g = 10 m/s2 Use g = 10 m/s2**A ball is thrown horizontally from A and passes through**B(d=20,h = -20) meters. • The travel time t to Point B is • t = 4 s • t = 1 s • t = 0.5 s • (D)t = 2 s • Use g = 10 m/s2 Use g = 10 m/s2**A ball is thrown horizontally from A and passes through**B(d=20,h = -20) meters at time t = 2s. • The start velocity v0 is • v0 = 40 m/s • v0 = 20 m/s • v0 = 10 m/s • (D) v0 = 5 m/s Use g = 10 m/s2**A ball is thrown horizontally from A and passes through**B(d=20,h = -20) meters at time t = 2s. • The start velocity v0 is • v0 = 40 m/s • v0 = 20 m/s • v0 = 10 m/s • (D) v0 = 5 m/s Use g = 10 m/s2**12.7 Normal and Tangential Coordinates**ut : unit tangent to the path un : unit normal to the path**Normal and Tangential Coordinates**Velocity Page 53**Fundamental Problem 12.27**The boat is traveling along the circular path with r = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is: • (A) constant • (B) 1 m/s2 • (C) 2 m/s2 • (D) not enough information • (E) 4 m/s2**Fundamental Problem 12.27**The boat is traveling along the circular path with r = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is: • (A) constant • (B) 1 m/s2 • (C) 2 m/s2 • (D) not enough information • (E) 4 m/s2**Polar Coordinates**• Point P moves on a counterclockwise circular path, with r =1m, q-dot(t) = 2 rad/s. The radial and tangential accelerations are: • (A) ar = 4m/s2 aq = 2 m/s2 • (B) ar = -4m/s2 aq = -2 m/s2 • (C) ar = -4m/s2 aq = 0 m/s2 • (D) ar = 0 m/s2 aq = 0 m/s2**Polar Coordinates**• Point P moves on a counterclockwise circular path, with r =1m, q-dot(t) = 2 rad/s. The radial and tangential accelerations are: • (A) ar = 4m/s2 aq = 2 m/s2 • (B) ar = -4m/s2 aq = -2 m/s2 • (C) ar = -4m/s2 aq = 0 m/s2 • (D) ar = 0 m/s2 aq = 0 m/s2**Point B moves radially outward from center C, with r-dot**=1m/s, q-dot(t) = 10 rad/s. At r=1m, the radial acceleration is: • (A) ar = 20 m/s2 • (B) ar = -20 m/s2 • (C) ar = 100 m/s2 • (D) ar = -100 m/s2**Point B moves radially outward from center C, with r-dot**=1m/s, q-dot(t) = 10 rad/s. At r=1m, the radial acceleration is: • (A) ar = 20 m/s2 • (B) ar = -20 m/s2 • (C) ar = 100 m/s2 • (D) ar = -100 m/s2**12.10 Relative (Constrained) Motion**We Solve Graphically (Vector Addition) vA vB vB/A**Example : Sailboat tacking against Northern Wind**2. Vector equation (1 scalar eqn. each in i- and j-direction) 500 150 i**Given:**• r(t) = 2+2*sin(q(t)), q_dot= constant • The radial velocity is • 2+2*cos(q(t ))*q-dot, • -2*cos(q(t))*q-dot • 2*cos(q(t))*q-dot • 2*cos(q(t)) • 2*q +2*cos(q(t ))*q-dot**Given:**• r(t) = 2+2*sin(q(t)), q_dot= constant • The radial velocity is • 2+2*cos(q(t ))*q-dot, • -2*cos(q(t))*q-dot • 2*cos(q(t))*q-dot • 2*cos(q(t)) • 2*q+2*cos(q(t ))*q-dot**Constrained Motion**vA is given as shown. Find vB Approach: Use rel. Velocity: vB = vA +vB/A (transl. + rot.)**The conveyor belt is moving to the left at v = 6 m/s. The**angular velocity of the drum (Radius = 150 mm) is • 6 m/s • 40 rad/s • -40 rad/s • 4 rad/s • none of the above**The conveyor belt is moving to the left at v = 6 m/s. The**angular velocity of the drum (Radius = 150 mm) is • 6 m/s • 40 rad/s • -40 rad/s • 4 rad/s • none of the above**Omit all constants!**• The rope length between points A and B is: • (A) xA – xB + xc • (B) xB – xA + 4xc • (C) xA – xB + 4xc • (D) xA + xB + 4xc**Omit all constants!**• The rope length between points A and B is: • (A) xA – xB + xc • (B) xB – xA + 4xc • (C) xA – xB + 4xc • (D) xA + xB + 4xc**Given: v0 = const.**• The vertical velocity component of point A (in y-direction) is • vA,y = v0*tan(q) • vA,y = v0*cot(q) • vA,y = v0*cos(q) • vA,y = 2*v0 q q**Vy/vx =cot q**q • Given: v0 = const. • The velocity of point A in vertical y-direction is • vA,y = v0*tan(q) • vA,y = v0*cot(q) • vA,y = v0*cos(q) • vA,y = 2*v0 • vA,y = v0/cos(q) q q**NEWTON'S LAW OF INERTIA A body, not acted on by any force,**remains in uniform motion. NEWTON'S LAW OF MOTION Moving an object with twice the mass will require twice the force. Force is proportional to the mass of an object and to the acceleration (the change in velocity). F=ma.**Dynamics**M1: up as positive: Fnet = T - m1*g = m1 a1 M2: down as positive. Fnet = F = m2*g - T = m2 a2 3. Constraint equation: a1 = a2 = a**Equations**From previous: T - m1*g = m1 a T = m1 g + m1 a Previous for Mass 2: m2*g - T = m2 a Insert above expr. for T m2 g - ( m1 g + m1 a ) = m2 a ( m2 - m1 ) g = ( m1 + m2 ) a ( m1 + m2 ) a = ( m2 - m1 ) g a = ( m2 - m1 ) g / ( m1 + m2 )**Rules**1. Free-Body Analysis, one for each mass 2. Constraint equation(s): Define connections. You should have as many equations as Unknowns. COUNT! 3. Algebra: Solve system of equations for all unknowns**Mass m rests on the 30 deg. Incline as shown. Step 1:**Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown. M*g*sinq*i -M*g*cosq*j M*g