Kinematics of Rigid Bodies
General Rotation of a Rigid Body
Learning Objectives:

Solve problems in twodimensional rigidbody dynamics, regardless of their kinematic characteristics, by equating the sum of the forces acting on the rigid body to the vectors ma and Ia. To effect this solution, construct appropriate freebody diagrams.
For rigid body rotation \(\theta\) \[\omega = \dfrac{d\theta}{dt}\] \[\alpha = \dfrac{d\omega}{dt}\] \[\alpha d\theta = \omega d\omega\]
Solved Example: 33101
Two objects are sitting on a rotating turntable. One is much further out from the axis of rotation. Which one has the larger angular velocity?
A. The one nearer the disk center
B. The one nearer the disk edge
C. They both have the same angular velocity
D. Cannot tell unless velocity and radii are given
Angular velocity depends upon angular displacement covered in unit time. Irrespective of the location of the objects, the angular displacement is same, hence there angular velocities are same. (Hoever, their linear velocities will be different.)
Correct Answer: C
Rotation of a Rigid Body under Constant Acceleration
Learning Objectives:

Draw, analyze and interpret angular displacementtime, angular velocitytime and angular accelerationtime graphs.

Solve simple problems involving uniform angular motion and uniformly accelerated motion by using appropriate equations.
\[\begin{aligned} \omega &= \omega_0 + \alpha t \\ \theta &= \theta_0 +\omega_0 t + \dfrac{1}{2} \alpha_0 t^2 \\ \omega^2 &= \omega_0^2 + 2\alpha_0 (\theta  \theta_0) \end{aligned}\]
where, \[\begin{aligned} \omega_0 &= \mathrm{Initial\ angular\ velocity}\\ \omega &= \mathrm{Final\ angular\ velocity}\\ t &= \mathrm{time}\\ \theta_0 &= \mathrm{Initial\ angular\ displacement}\\ \theta &= \mathrm{angular\ displacement}\\ \alpha &= \mathrm{angular\ acceleration}\\ \end{aligned}\]
Solved Example: 33201
A bike has a front wheel of radius 0.25 m. It starts from rest, and accelerates down a hill with linear acceleration a= 2.0 m/s$^2$. At t= 1.0 sec, what is the angular velocity $\omega$ of the bike wheel?
A. 1.0 rad/s
B. 4.0 rad/sec
C. 8.0 rad/s
D. 8 $\pi$ rad/s
\[\alpha =a/r = \dfrac{2}{0.25} = 8.0\ rad/s^2\] \[\omega = \omega_0 + \alpha t =0 + (8)(1) = 8 rad/s\]
Correct Answer: C
Solved Example: 33202
A cylinder (r = 0.14 m, I$_{cm}$ = 2.4 $\times$ 10$^{2}$ kg.m$^2$, M = 1.5 kg) starts at rest and rolls without slipping down a plane with an inclination angle of $\theta$ =15$^\circ$. Find the time it takes to travel 1.4 m.
A. 0.7 sec
B. 1.4 sec
C. 2.7 sec
D. 2.9 sec
Use Newton's second law, \[\sum F_{x}=mg\sin\thetaF_{FR}=ma\] Torque about center of mass \[\sum \tau=rF_{FR}=I\alpha=\dfrac {Ia}{r}\] \[F_{FR}=\dfrac {Ia}{r^{2}}\] \[mg\sin\thetaI\dfrac {a}{r^{2}}=ma\] \[a=\dfrac {r^{2}mg\sin\theta}{I+mr^{2}}=1.4m/s^{2}\] Now using, \begin{align*} x&=x_{0}+v_{0}t+\dfrac {1}{2}at^{2}\\ 1.4&=0+0+\dfrac {1}{2}\left( 1.4\right) t^{2}\\ t&=1.4s \end{align*}Correct Answer: B
Rotation of a Rigid Body under Variable Acceleration
Learning Objectives:

Integrate angular acceleration function to find angular velocity and angular displacement.
If a rigid body is under variable acceleration \[\alpha = f(t)\] then we have to take the acceleration function, and integrate it twice to get angular displacement function.
If a rigid body is under variable acceleration \[\alpha = f(\omega)\] or \[\alpha = f(\theta)\] then we have to separate the variables using \[\alpha d\theta = \omega d\omega\] and integrate the variable separable differential equation twice to get angular displacement function.
Solved Example: 33301
A wheel starts is spinning at 27 rad/s but is slowing with an angular acceleration that has a magnitude given by $\alpha$ (t)= 3t$^2$. It stops in a time of:
A. 2 sec
B. 3 sec
C. 3.5 sec
D. 9 sec
\[\alpha (t) =  3t^2\] \begin{align*} \omega &= \omega_0 + \int \alpha (t) dt\\ &= 27  t^3 \end{align*} When the wheel stops, \begin{align*} 0 &= 27 t^3\\ t &= 3\ sec \end{align*}
Correct Answer: B
Instantaneous Center of Rotation
Learning Objectives:
 State Methods for determining the velocity of a point on a link.
 Define Instantaneous Centre of Rotation.
 State properties of the instantaneous centre.
 Calculate velocity of a point on a link by instantaneous centre method.
It denotes the center of rotation of a body at an instant in time. The center of rotation doesn’t necessarily have to lie within the link itself. The instantaneous center between bodies 1 and 2 is defined as \(I_{12}\).

It is a point in one body about which some other body is permanently or instantaneously rotating about.

It is a point common to two bodies where the velocity of the two bodies are the same. i.e., a point where there is zero relative velocity between the two bodies.
To find the instant center for a single body when the velocities of two points are known we take advantage of the fact that the linear velocities of all points in a rotating body are perpendicular to their radii of rotation. The two points A and B along with their known velocity vectors \(v_A\) and \(v_B\) determine the I.C. If the two velocity vectors are equal (i.e., same magnitude and direction) then the I.C. is a point at infinity, and the body’s motion is in pure translation.
I.C. for a Rigid Body Rotating in a Plane:
If velocities of two points on a rigid body are known, then draw lines perpendicular to the direction of these velocities. The intersection point of these lines is the instantaneous center of rotation.
I.C. for a Sliding Body:
It lies on a line perpendicular to the motion at infinite distance. (It may be on either side).
I.C. for a Rolling Body:
The instantaneous center for a rolling body occurs at the point of contact between the two bodies.
Solved Example: 33401
A wheel is rolling without slipping on a plane surface with the center of velocity V. What will be the velocity at the point of contact?
A. Velocity V perpendicular to the surface
B. Velocity V in the direction of motion
C. V in the opposite direction of motion
D. Zero
For a rotating wheel (rotation without slipping) the instantaneous center is the point of contact. Hence, its instantaneous velocity is zero.
Correct Answer: D
Solved Example: 33402
Which is the false statement about the properties of instantaneous centre?
A. At the instantaneous centre of rotation, one rigid link rotates instantaneously relative to another for the configuration of mechanism considered
B. The two rigid links have no linear velocities relative to each other at the instantaneous centre
C. The two rigid links which have no linear velocity relative to each other at this centre have the same linear velocity to the third rigid link
D. The double centre can be denoted either by O$_{21}$ or O$_{12}$, but proper selection should be made
The following properties of the instantaneous centre are important:
A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.
The two rigid links have no linear velocity relative to each other at the instantaneous centre. At this point (i.e. instantaneous centre), the two rigid links have the same linear velocity relative to the third rigid link. In other words, the velocity of the instantaneous centre relative to any third rigid link will be same whether the instantaneous centre is regarded as a point on the first rigid link or on the second rigid link.
Correct Answer: D
Solved Example: 33403
Instantaneous center of rotation of a link with respect to its adjoining link in a four bar mechanism lies on:
A. Right side pivot of this link
B. Left side pivot of this link
C. A point obtained by intersection on extending adjoining links
D. None of the mentioned
Correct Answer: C
Solved Example: 33404
The total number of instantaneous centers for a mechanism of n links is:
A. $N = \dfrac{n(n  1)}{2}$
B. n
C. n  1
D. n(n  1)
The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, $N = \dfrac{n(n  1)}{2}$
Correct Answer: A
Solved Example: 33405
A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the xaxis, and the body moves such that velocities of points P and R are V$_P$ and V$_R$ , in the x and y directions, respectively. The magnitude of the angular velocity of the body is:
A. 2V$_R$
B. 2V$_P$
C. $\dfrac{V_R}{\sqrt{3}}$
D. $\dfrac{V_P}{\sqrt{3}}$
This problem can be solved using I.C. (Instantaneous Center) concept. Instantaneous velocities are always perpendicular to I.C. Draw perpendiculars from both velocity vectors, their intersection point is I.C. as shown in the figure. Consider the velocity V$_R$. Distance from I.C. to R = $\dfrac{1}{2}$ as QR = 1 unit (given)
\begin{align*} V &= r \omega \\ V_R &= \dfrac{1}{2} \omega\\ \omega &= 2 V_R \end{align*}Correct Answer: A
Solved Example: 33406
When a slider moves on a fixed link having curved surface, the instantaneous center lies on: (MP Sub Engg Mechanical July 2017  Shift II)
A. Center of curvature of curvilinear path
B. Central axis of slider
C. At infinity
D. Point of contact between links
When a slider moves on a fixed link having flat surface, it has instantenous center at infinity. However, if the sliding surface has curvature, the slider is effectively undergoing a circular motion. The center of this motion is the center of curvature of the curved surface.
Correct Answer: A
Solved Example: 33407
A circular object of radius 'r' rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is: (GATE ME 2014)
A. Zero
B. V in the direction of motion
C. V opposite to the direction of motion
D. V vertically upward from the floor
The key phrase here is 'at the point of contact', where the velocity will always be zero, unless the objects are skidding.
Correct Answer: A
Solved Example: 33408
The number of instantaneous centres of rotation for a 10link kinematic chain is: (ESE Mechanical 2014)
A. 36
B. 90
C. 120
D. 45
If N is the number of instantaneous centres and n is the number of links then: \begin{align*} N &= \dfrac{{n\left( {n  1} \right)}}{2}\\ N &= \dfrac{10(101)}{2}\\ N &= 45 \end{align*}
Correct Answer: D
Kennedys Rule
Learning Objectives:

Calculate number of instantaneous centres in a mechanism.

Identify types and locations of instantaneous centres.

Analyze the location of ICR to solve for system containing multiple links.

Calculate the velocities at different location of the mechanism
When three bodies move relative to one another, they have three instantaneous centers, all of which lie on the same straight line.
Solved Example: 33501
According to Kennedy's theorem, if three bodies have plane motions, their instantaneous centers lie on:
A. A triangle
B. A point
C. Two lines
D. A straight line
Let's assume that the three bodies are 1, 2 and 3.
There will be total $\dfrac{3(31)}{2}= 3$ Instantaneous centers of rotations. Let's call them $I_{12}, I_{23}$ and $I_{13}$ respectively.
Then, according to Kennedy's theorem, $I_{12}, I_{23}, I_{13}$ will lie on a straight line.
Correct Answer: D