# ENGINEERING MECHANICS CHAPTER 7

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ENGINEERING MECHANICS CHAPTER 7
School of Engineering ENGINEERING MECHANICS CHAPTER 7 Kinematics of Linear & Rotational Motion Click on VIEW and select SLIDE SHOW to view the presentation.

6.1) Introduction To Kinematics of Linear Motion
Kinematics is the analysis of the geometry of motion without considering the forces causing the motion. It involves quantities like displacement (m), velocity (m/s), acceleration (m/s2) and time (s). 6.1.1 Displacement Displacement is defined as the location of the body with respect to a reference point. It specifies the distance and the direction from the reference point. The SI unit for displacement is m.

Displacement-time graph is a plot of the displacement against time.
Example: An object is thrown vertically upward from the top of a building. The starting location is taken as the reference point. Its displacement from starting location is as shown in the figure below. Negative displacement s t Maximum height Zero displacement Reference point

6.1.2 Velocity Velocity is the rate of change of displacement with respect to time. The SI unit for velocity is m/s. v = ds/dt Velocity-time Graph The velocity-time graph is useful to visualize the motion. Using the same example above, the plot of the velocity with respect to time is as shown in the figure below. During the upward journey, the velocity is positive. It decreases to zero when the body reaches the maximum height. The velocity becomes negative on the downward journey.

Zero velocity at maximum height
Positive velocity Negative velocity Acceleration Acceleration is the rate of change of velocity with respect to time. The SI unit for acceleration is m/s2. a = dv/dt Using the same example as above, the acceleration will be a constant and is negative for the whole journey. The magnitude of the acceleration is 9.81 m/s2 for free falling motion.

Uniform motion A motion is called uniform motion when the acceleration is constant. The three equations of motion with constant acceleration are: v = vi + at v2 = vi2 + 2as s = vit + ½at2 where v is the final velocity vi is the initial velocity a is the acceleration s is the displacement t is the time

Example 6.1 A projectile is ejected vertically from ground level upward with an initial velocity of 30 m/s. Determine: i) its velocity after 2 s, ii) the height above the ground at this time, iii) the time taken to reach its maximum height, iv) the total time taken for it to hit the ground again. Solution: i) Take upward as positive direction. vi = 30 m/s a = m/s2 t = 2 s i.e v = vi + at = 30 + (- 9.81) x 2 = m/s

ii) s = vit + ½at2 = 30 x 2 + ½ x (-9.81) x 22 = m iii) At maximum height, velocity is zero. v = vi + at 0 = 30 + (-9.81) x t t = s iv) s = vit + ½at2 0 = 30 x t + ½ x (-9.81) x t2 t = 0 or s

Non-uniform Motion A moving body with non-uniform motion has different accelerations (and/or decelerations) during the whole motion. The motion can be divided into several segments with constant acceleration for each segment. The three equations for constant acceleration can then be applied for each segment where the motion is uniform.

i) the velocity at point B, ii) the distance from A to B,
Example 6.2 An MRT train travels between two stations A and D. It starts with an acceleration of 1.2 m/s2 for 18 s until it reaches point B. The velocity remains constant over a distance of 2000 m from point B to C, and then it decelerates uniformly at 2.4 m/s2 until it stops at D. Determine: i) the velocity at point B, ii) the distance from A to B, iii) the distance C to D, iv) the total time and distance. D C B A v t Solution: Velocity - Time Graph

i) From A to B, vi = 0 m/s a = 1.2 m/s2 t = 18 s, v = vi + at
D C B A v t - 2.4 m/s2 1.2 m/s2 18 s t2 t1 i) From A to B, vi = 0 m/s a = 1.2 m/s2 t = 18 s, v = vi + at = x 18 = 21.6 m/s ii) From A to B, s = vit + ½at2 = 0 x 18 + ½ x 1.2 x 182 = m iii) From C to D, vi = 21.6 m/s Also, v = 0 and a = m/s2 v2 = vi2 + 2as 0 = x (-2.4) x s s = 97.2 m

iv) Total distance, sT = 194.4 + 2000 + 97.2 = 2291.6 m
B A v t - 2.4 m/s2 1.2 m/s2 18 s t2 t1 iv) Total distance, sT = = m From B to C, vi= 21.6 m/s a = 0, s = 2000 m s = vit + ½at2 2000 = 21.6 x t + 0 t = s (t1) From C to D, v = vi + at 0 = (-2.5) x t t = 8.64 s (t2) Total time, tT = = s (Ans)

6.2) Introduction To Kinematics of Rotational Motion
Kinematics of rotational motion involves quantities like angular displacement, angular velocity, angular acceleration and time. Angular Displacement Angular displacement is defined as the angle and the direction through which a body turns. The SI unit for angular displacement is radian (rad). ( Note:- 1 revolution = 360 degrees = 2 radians ) Angular Velocity Angular velocity is the rate of change of angular displacement with respect to time. The angular velocity can be written in mathematical form as:  = d/dt The SI unit for angular velocity is rad/s.

Angular acceleration is the rate of change of
angular velocity with respect to time. The angular acceleration can be written in mathematical form as:  = d/dt The SI unit for  is rad/s2. Uniform motion A motion is called uniform when the angular acceleration is constant. The three equations of rotational motion with constant acceleration are: Note the similarities of these equations with those of linear motion in above. Where  is the final velocity i is the initial velocity  is the acceleration  is the displacement t is the time  = i + t 2 = i2 + 2  = it + ½t2

6.2.5 Relationship Between Linear and Rotational motion
The equations for the relationship between linear and rotational motion are: s = r v = r ( Divide s = r by t ) a = r ( Divide v = r by t ) where r is the radius of rotation The linear velocity and acceleration are also the tangential velocity and acceleration respectively for a point on the circle of rotation. ( Note: v = vt and a = at ) s = r

Example 6.3 A propeller fan used in a cooling tower comes to rest with uniform deceleration from a speed of 600 rpm. It turns through 15 rev while stopping. Determine the initial blade tip speed if the blade is 2 m long, the deceleration, the time taken to stop. Solution i.  i = 600 rpm = (600 x 2) / 60 = rad/s v = r = 2 x 62.83 = m/s

ii  = 15 rev = 15 x 2 = 30 rad  = 0 , i =62.83 rad/s since 2 = i2 + 2 0 = x  x 30  = rad/s2 iii  = i + t 0 = ( )t t = 3 s

the number of revolutions made by the pulley in 3 s,
Example 6.4 A compound pulley has an outer radius of 1.25 m and an inner radius of 0.75 m. Load A and load B are connected by cords to the pulley at the outer radius and inner radius respectively. Load A has a constant acceleration of 2.5 m/s2 and an initial velocity of 3.75 m/s, both in the upward direction. Determine the number of revolutions made by the pulley in 3 s, the velocity of load B in this time, the distance moved by load B. A B r = 0.75 m r = 1.25 m vi = 3.75 m/s a = 2.5 m/s2

i) vi = ri 3.75 = 1.25i i = 3 rad/s a = r 2.5 = 1.25  = 2 rad/s2
B r = 0.75 m r = 1.25 m vi = 3.75 m/s a = 2.5 m/s2 At t = 3 s,  = it + ½t2 = 3 x 3 + ½ x 2 x 32 = 18 rad Number of revolutions, N = 18 / 2 = 2.87 rev

iii) Distance moved by load B, s = r = 0.75 x 18 = 13.5 m
ii)  = i + t = x 3 = 9 rad/s v = r = 0.75 x 9 = 6.75 m/s A B r = 0.75 m r = 1.25 m vi = 3.75 m/s a = 2.5 m/s2 iii) Distance moved by load B, s = r = 0.75 x 18 = 13.5 m

6.2.6 Normal or Centripetal Acceleration
For a body moving in a circular motion, it has an acceleration an, towards the center of the circle with a magnitude given by, at (Tangential Acceleration) (Normal or Centripetal Acceleration) an aT (Total Acceleration) an = v or an = v2/r (Since  = v/r) or an = r (Since v = r) ( Note:- Total acceleration aT2 = an2 + at2 )

Example 6.5 Disk A and disk B are on the same plane. Disk A starts from rest and rotates with a constant angular acceleration of 2 rad/s2. It is in contact with disk B and no slipping occurs between them. Determine the angular velocity and angular acceleration of B just after A turns 10 revolutions. What is the total acceleration of the point P in relation to disk B? A B P 2m 1.5m

Velocity of point P on disk A is vA = rAA = 2 x 15.85 = 31.7 m/s
Solution A2 = Ai2 + 2A = x 2 x (10 x 2) A = rad/s Velocity of point P on disk A is vA = rAA = 2 x 15.85 = 31.7 m/s A B P 2m 1.5m Since there is no slipping, vB = vA = 31.7 m/s Then vB = rBB 31.7 = 1.5B B = rad/s Acceleration of the point P on disk A is aA = rAA = 2 x 2 = 4 m/s2

End of Chapter 6 Since there is no slipping, aB = aA = 4 m/s2
Then aB = rBB 4 = 1.5B B = 2.67 rad/s2 A B P 2m 1.5m For point P in relation to disk B, at = aB = 4 m/s2 an = r2 = 1.5 x = m/s2 aT2 = at2 + an2 = aT = m/s2 End of Chapter 6