Kinematics of Particles and Rigid Bodies Notes FE Sem 1 | MU
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Kinematics answers where a body is, how fast it moves and how its speed and direction change, without asking which forces cause the motion. The first part of this module treats particles: motion along a line when acceleration varies with time, velocity or position, and motion along curved paths described by rectangular components or by tangential and normal components. The second part introduces general plane motion of rigid bodies and the instantaneous centre of rotation (ICR) method for links such as a sliding ladder, a rolling wheel and a crank and connecting rod. These tools are the basis for kinetics in Module 5 and for machine and robot motion later.
Summary
Motion of a particle with variable acceleration: velocity and acceleration are derivatives of position, v = dx/dt and a = dv/dt = d^2x/dt^2 = v dv/dx. When acceleration is not constant, the equations of uniformly accelerated motion are invalid and we integrate instead. If a = f(t), integrate dv = f(t) dt; if a = f(v), separate dv/f(v) = dt or v dv/f(v) = dx; if a = f(x), use v dv = f(x) dx. Initial conditions fix the constants. Distance travelled differs from displacement when the velocity changes sign, so find the instants when v = 0 and add the distances of each interval.
Motion along a plane curved path: in curvilinear motion the position vector r = x i + y j changes in magnitude and direction. The velocity is always tangent to the path, and acceleration generally has a component towards the centre of curvature even at constant speed. The path equation y = f(x) is found by eliminating time between x(t) and y(t).
Velocity and acceleration in rectangular components: vx = dx/dt, vy = dy/dt, ax = dvx/dt, ay = dvy/dt. Magnitudes are v = square root of (vx^2 + vy^2) and a = square root of (ax^2 + ay^2), and the direction of v gives the tangent to the path. Projectile motion is the classic case: ax = 0, ay = -g.
Tangential and normal components of acceleration: the tangential component at = dv/dt changes the speed and acts along the tangent; the normal component an = v^2/rho changes the direction and acts towards the centre of curvature, where rho is the radius of curvature. Total acceleration a = square root of (at^2 + an^2). From rectangular components, at = (vx ax + vy ay)/v and an = |vx ay - vy ax|/v, and rho = v^2/an; for a path y = f(x), rho = [1 + (dy/dx)^2]^(3/2) / |d^2y/dx^2|.
Introduction to general plane motion: a rigid body in plane motion can undergo pure translation (every line stays parallel to itself), rotation about a fixed axis (points move on circles, v = r omega, at = r alpha, an = r omega^2) or general plane motion, which is a combination of translation and rotation, such as a rolling wheel or a connecting rod. All points of a rigid body have the same angular velocity at an instant.
Instantaneous centre of rotation (ICR): at any instant a body in general plane motion can be treated as rotating purely about a point, the ICR, which has zero velocity at that instant. It lies at the intersection of the perpendiculars drawn to the velocity directions of two points of the body; if the velocities are parallel and perpendicular to the line joining the points, it lies where the line joining the velocity vector tips meets that line. Then omega = vA/IA and the velocity of any point P is vP = omega x IP, perpendicular to IP. For a wheel rolling without slipping the ICR is the contact point, so v_centre = r omega. The ICR changes position from instant to instant and has non-zero acceleration, so it is used for velocities only. The syllabus limits problems to mechanisms of up to two links and excludes the relative velocity method.
Mind map
Kinematics of Particle and Rigid Bodies
- Variable acceleration
- a = f(t)
- a = f(v)
- a = f(x)
- v dv/dx form
- Distance vs displacement
- Curvilinear motion
- Position vector
- Path equation
- Projectile as example
- Rectangular components
- vx, vy
- ax, ay
- Magnitude and direction
- Normal-tangential
- at = dv/dt
- an = v^2/rho
- Radius of curvature
- Rigid body motion
- Translation
- Fixed-axis rotation
- General plane motion
- ICR method
- Locating the ICR
- omega = v/r
- Ladder, rolling wheel
- Crank and connecting rod
Finding velocities by the instantaneous centre of rotation method
- Draw the link to scale or with dimensions at the given instant and identify two points A and B whose velocity directions are known.
- Draw lines through A and B perpendicular to their velocity directions; their intersection is the ICR, I. (For a rolling wheel, I is the contact point.)
- Find distances IA and IB using geometry or trigonometry (sine rule).
- Compute the angular velocity of the link: omega = vA / IA.
- Find the velocity of the other point: vB = omega x IB, directed perpendicular to IB in the sense of rotation.
- For any other point P of the link, vP = omega x IP perpendicular to IP.
- For a two-link mechanism, first find the velocity of the common joint from the driving link, then apply steps 2 to 6 to the second link.
Important definitions
| Term | Meaning |
|---|---|
| Kinematics | The study of motion of bodies in terms of displacement, velocity and acceleration without considering the forces causing it. |
| Displacement | The change in position of a particle, a vector from initial to final position. |
| Distance travelled | The total length of path covered, a scalar that is always positive. |
| Variable acceleration | Acceleration that changes with time, velocity or position, requiring calculus instead of the constant-acceleration equations. |
| Curvilinear motion | Motion of a particle along a curved path. |
| Tangential acceleration | Component of acceleration along the tangent to the path; it changes the magnitude of velocity. |
| Normal acceleration | Component of acceleration along the normal towards the centre of curvature; it changes the direction of velocity, equal to v^2/rho. |
| Radius of curvature | The radius of the circle that best fits the path at a point. |
| Translation | Motion of a rigid body in which every line in the body remains parallel to its original position. |
| General plane motion | Plane motion of a rigid body that is a combination of translation and rotation. |
| Instantaneous centre of rotation | The point, in the body or its extension, which has zero velocity at a given instant, so the body appears to rotate purely about it at that instant. |
| Angular velocity | Rate of change of angular position of a rigid body, omega = d theta/dt, in rad/s; same for all points of the body. |
Formulas
| Formula | Meaning |
|---|---|
| \( v = \frac{dx}{dt},\quad a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = v\frac{dv}{dx} \) | Basic relations for rectilinear motion; use v dv/dx when acceleration is given in terms of position. |
| \( \int_{v_0}^{v} dv = \int_0^t f(t)\,dt \) | Integration when a = f(t). |
| \( \int_{v_0}^{v} \frac{dv}{f(v)} = t,\quad \int_{v_0}^{v} \frac{v\,dv}{f(v)} = x - x_0 \) | Integration when a = f(v), for time and for distance respectively. |
| \( \int_{v_0}^{v} v\,dv = \int_{x_0}^{x} f(x)\,dx \) | Integration when a = f(x). |
| \( v = \sqrt{v_x^2 + v_y^2},\quad a = \sqrt{a_x^2 + a_y^2} \) | Magnitudes from rectangular components vx = dx/dt, vy = dy/dt, ax = dvx/dt, ay = dvy/dt. |
| \( a_t = \frac{dv}{dt},\quad a_n = \frac{v^2}{\rho},\quad a = \sqrt{a_t^2 + a_n^2} \) | Tangential and normal components; rho is the radius of curvature (m). |
| \( a_t = \frac{v_x a_x + v_y a_y}{v},\quad a_n = \frac{|v_x a_y - v_y a_x|}{v} \) | Tangential and normal acceleration obtained from rectangular components. |
| \( \rho = \frac{\left[1 + (dy/dx)^2\right]^{3/2}}{|d^2y/dx^2|} \) | Radius of curvature of a path y = f(x). |
| \( v = r\omega,\quad a_t = r\alpha,\quad a_n = r\omega^2 \) | Point at radius r on a body rotating about a fixed axis. |
| \( \omega = \frac{v_A}{IA},\quad v_P = \omega \cdot IP \) | ICR method: I is the instantaneous centre; velocity of any point is perpendicular to the line joining it to I. |
Key points to remember
- Constant-acceleration equations (v = u + at etc.) must not be used when acceleration varies.
- Always check whether velocity becomes zero within the time interval before computing distance travelled.
- Velocity in curvilinear motion is always tangent to the path.
- A particle moving at constant speed on a curve still accelerates because of the normal component v^2/rho.
- At the top of a projectile path the velocity is horizontal, so the whole of g is normal acceleration and rho = v^2/g.
- All points of a rigid body have the same angular velocity at an instant, but different linear velocities.
- The ICR has zero velocity but generally non-zero acceleration, so use it only for velocity analysis.
- For rolling without slipping the contact point is the ICR; the top of the wheel moves at twice the centre velocity.
- In a slider-crank mechanism the ICR of the connecting rod lies on the extended crank line and on the perpendicular to the slider path through the slider.
- Relative velocity method is excluded; use the ICR method only, for up to two-link mechanisms.
Solved examples
Example 1. The position of a particle moving along a straight line is x = t^3 - 6t^2 + 9t + 5 m, t in seconds. Find (a) when the velocity is zero, (b) the position and acceleration at t = 4 s, (c) the total distance travelled in the first 4 s.
v = dx/dt = 3t^2 - 12t + 9 = 3(t - 1)(t - 3); a = dv/dt = 6t - 12. (a) v = 0 at t = 1 s and t = 3 s. (b) x(4) = 64 - 96 + 36 + 5 = 9 m; a(4) = 24 - 12 = 12 m/s^2. (c) x(0) = 5 m, x(1) = 1 - 6 + 9 + 5 = 9 m, x(3) = 27 - 54 + 27 + 5 = 5 m, x(4) = 9 m. Distance = |9 - 5| + |5 - 9| + |9 - 5| = 4 + 4 + 4 = 12 m (displacement is only 4 m).
Example 2. A body moving with an initial velocity of 20 m/s is retarded by a = -0.4v m/s^2. Find the time taken and the distance covered while its velocity drops to 5 m/s.
Time: dv/dt = -0.4v, so dv/v = -0.4 dt. Integrating from 20 to 5: ln(5/20) = -0.4t, t = ln 4/0.4 = 3.466 s. Distance: v dv/dx = -0.4v, so dv = -0.4 dx. Integrating: 5 - 20 = -0.4x, x = 15/0.4 = 37.5 m.
Example 3. The acceleration of a particle is a = -4x m/s^2, where x is in metres. At x = 0 the velocity is 8 m/s. Find the velocity at x = 2 m and the maximum displacement.
v dv = -4x dx. Integrating from (x = 0, v = 8): (v^2 - 64)/2 = -2x^2, so v^2 = 64 - 4x^2. At x = 2 m, v = square root of (64 - 16) = 6.93 m/s. Maximum displacement when v = 0: 4x^2 = 64, x = 4 m.
Example 4. A particle moves in the x-y plane with x = 3t^2 and y = t^3 - 2t (metres, seconds). At t = 2 s find the velocity, acceleration, tangential and normal components of acceleration and the radius of curvature.
vx = 6t = 12 m/s, vy = 3t^2 - 2 = 10 m/s; v = square root of (144 + 100) = 15.62 m/s at arctan(10/12) = 39.81 degrees to x. ax = 6 m/s^2, ay = 6t = 12 m/s^2; a = square root of (36 + 144) = 13.42 m/s^2. at = (vx ax + vy ay)/v = (72 + 120)/15.62 = 12.29 m/s^2. an = |vx ay - vy ax|/v = |144 - 60|/15.62 = 5.38 m/s^2. Check: 12.29^2 + 5.38^2 = 180 = 13.42^2. rho = v^2/an = 244/5.38 = 45.37 m.
Example 5. A car moves on a curved road of radius 200 m at 72 km/h and its speed is decreasing at 1.5 m/s^2. Find its total acceleration.
v = 72 x 1000/3600 = 20 m/s. an = v^2/rho = 400/200 = 2 m/s^2 towards the centre. at = -1.5 m/s^2 (opposite to motion). a = square root of (2^2 + 1.5^2) = 2.5 m/s^2, inclined at arctan(2/1.5) = 53.13 degrees to the tangent.
Example 6. A projectile is fired at 40 m/s at 30 degrees to the horizontal. Find the radius of curvature of its path at the highest point.
At the highest point the vertical velocity is zero, so v = 40 cos 30 = 34.64 m/s, horizontal. The only acceleration is g = 9.81 m/s^2 downward, which is normal to the path there. rho = v^2/an = 34.64^2/9.81 = 1200/9.81 = 122.32 m.
Example 7. A ladder AB 5 m long slides with end A on the floor and end B against a vertical wall. When A is 3 m from the wall it moves away at 2 m/s. Find the angular velocity of the ladder, the velocity of B and of the midpoint G.
B is 4 m above the floor (3-4-5 triangle). vA is horizontal, so the perpendicular at A is vertical; vB is vertical, so the perpendicular at B is horizontal. They meet at I, 3 m from the wall and 4 m above the floor. IA = 4 m, IB = 3 m, IG = half the diagonal of the 3 x 4 rectangle = 2.5 m. omega = vA/IA = 2/4 = 0.5 rad/s. vB = omega x IB = 0.5 x 3 = 1.5 m/s downward. vG = 0.5 x 2.5 = 1.25 m/s, perpendicular to IG.
Example 8. In a slider-crank mechanism the crank OA = 0.2 m rotates at 10 rad/s and the connecting rod AB = 0.6 m; slider B moves along the horizontal line through O. For a crank angle of 45 degrees, find the angular velocity of the rod and the velocity of the slider using the ICR method.
vA = 0.2 x 10 = 2 m/s perpendicular to OA. Coordinates: A = (0.1414, 0.1414) m. Horizontal projection of AB = square root of (0.6^2 - 0.1414^2) = 0.5831 m, so OB = 0.7245 m. The ICR of AB lies on OA produced (perpendicular to vA) and on the vertical through B (perpendicular to vB): I = (0.7245, 0.7245). IA = OI - OA = 0.7245 x 1.4142 - 0.2 = 0.8246 m; IB = 0.7245 m. omega_AB = vA/IA = 2/0.8246 = 2.43 rad/s. vB = omega_AB x IB = 2.43 x 0.7245 = 1.76 m/s along the line of stroke.
Important questions and answers
Q1. Explain the instantaneous centre of rotation and how it is located. (4 marks)
For a body in general plane motion, the ICR is the point which has zero velocity at that instant; the motion then appears as pure rotation about it. To locate it, draw perpendiculars to the velocity directions of two points of the body; they intersect at the ICR. If two velocities are parallel and perpendicular to the line joining the points, the ICR lies on that line where it meets the line through the tips of the velocity vectors. If the velocities are parallel and equal, the ICR is at infinity and the body is in translation. For rolling without slipping it is the contact point.
Q2. Derive expressions for tangential and normal components of acceleration. (5 marks)
Let a particle move on a curve with speed v and unit tangent e_t. Velocity v = v e_t. Differentiating, a = (dv/dt) e_t + v (de_t/dt). As the particle moves ds along the path, the tangent turns by d theta = ds/rho, and de_t = d theta e_n, where e_n is the unit normal towards the centre of curvature. So de_t/dt = (d theta/dt) e_n = (1/rho)(ds/dt) e_n = (v/rho) e_n. Hence a = (dv/dt) e_t + (v^2/rho) e_n: tangential component at = dv/dt, normal component an = v^2/rho.
Q3. Differentiate between rectilinear and curvilinear motion, and between translation and rotation. (3 marks)
Rectilinear motion is along a straight line, so velocity and acceleration are along that line. Curvilinear motion is along a curved path and has normal acceleration. Translation is rigid body motion in which every line remains parallel to its original position, all points having the same velocity. Rotation about a fixed axis has all points moving on circles about the axis with v = r omega.
Q4. The velocity of a particle is v = 2t^2 - 3t m/s. If x = 2 m at t = 0, find the position and acceleration at t = 3 s. (4 marks)
x = integral of v dt = 2t^3/3 - 3t^2/2 + C; at t = 0, x = 2, so C = 2. x(3) = 18 - 13.5 + 2 = 6.5 m. a = dv/dt = 4t - 3, a(3) = 9 m/s^2.
Q5. A wheel of radius 0.5 m rolls without slipping on a horizontal road with its centre moving at 5 m/s. Find the velocities of the top point and of a point on the rim level with the centre. (4 marks)
ICR is the contact point. omega = v_centre/r = 5/0.5 = 10 rad/s. Top point: distance from ICR = 2r = 1 m, v = 10 m/s horizontal, forward. Rim point level with the centre: distance from ICR = square root of (0.5^2 + 0.5^2) = 0.707 m, v = 10 x 0.707 = 7.07 m/s, at 45 degrees to the horizontal (perpendicular to the line from the contact point).
Q6. A particle moves along a path y = x^2/20 (m) with a constant speed of 10 m/s. Find the normal acceleration at x = 10 m. (4 marks)
dy/dx = x/10 = 1 at x = 10; d^2y/dx^2 = 1/10. rho = (1 + 1)^(3/2)/(0.1) = 2.828/0.1 = 28.28 m. Since speed is constant, at = 0 and an = v^2/rho = 100/28.28 = 3.54 m/s^2, which is the total acceleration.
Q7. Explain the method of solving problems when acceleration is a function of velocity. (3 marks)
When a = f(v), write a = dv/dt to get time: dt = dv/f(v), integrate between the initial and final velocities. To get distance, write a = v dv/dx: dx = v dv/f(v), and integrate. Initial conditions supply the limits. Example: for a = -kv, v = v0 e^(-kt) and x = (v0 - v)/k.
Q8. State the characteristics of the instantaneous centre. (3 marks)
1. It is the point of zero velocity at the given instant. 2. Its position changes from instant to instant (it traces centrodes). 3. All points of the body move perpendicular to the lines joining them to the ICR, with speed proportional to that distance. 4. Its acceleration is generally not zero, so it cannot be used for accelerations. 5. It may lie inside or outside the body; for translation it is at infinity.
Q9. A point on a rotating disc of radius 0.3 m has angular velocity 4 rad/s and angular acceleration 2 rad/s^2. Find the total acceleration of a point on the rim. (3 marks)
at = r alpha = 0.3 x 2 = 0.6 m/s^2; an = r omega^2 = 0.3 x 16 = 4.8 m/s^2. a = square root of (0.6^2 + 4.8^2) = 4.84 m/s^2.
Exam tips
- Write the given relation and the substitution used (dv/dt or v dv/dx) clearly before integrating.
- Apply limits rather than leaving constants of integration undetermined.
- For distance travelled, find times when v = 0 and tabulate positions; examiners deliberately set this trap.
- Draw the ICR construction clearly with perpendiculars, distances and the sense of rotation.
- State velocity directions in the final answer (perpendicular to the ICR line), not only magnitudes.
- Convert km/h to m/s before using v^2/rho.
FAQs
When do I use v dv/dx instead of dv/dt?
Use v dv/dx when acceleration is given as a function of position or when you need a relation between velocity and distance without time.
Can the ICR be used to find acceleration?
No. The ICR has zero velocity but not zero acceleration, so it is used only for velocity analysis.
Which ICR problems are included?
Mechanisms up to two links, such as a sliding ladder or rod, a rolling wheel, and crank with connecting rod. The relative velocity method is excluded.
Is projectile motion examined here?
Basic projectile formulas are prerequisites, but projectile motion is used as an example of curvilinear motion, for example the radius of curvature at a point of the trajectory.
These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Engineering Mechanics) and were prepared with AI help and checked.
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