9.3 Relate velocities of points on a planar rigid body | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Introduction to Rigid-Body Dynamics

EN PH 131 study topic 9.3

A rigid body can translate and rotate at the same time. Its points therefore need not have the same velocity, even though their relative positions stay fixed. In this lesson, the system is one planar rigid body, observed from a stationary ground frame. We use fixed horizontal and vertical axes, with positive

x

to the right, positive

y

upward, and positive angular velocity counterclockwise. The key task is to connect the velocity of one point to that of another point on the same body. No force or acceleration analysis is needed to make this connection.

What you will learn

Define a planar rigid-body system, observer, reference frame, and coordinate directions.
Use the relative-velocity equation to relate the velocities of two points on a rigid body.
Determine velocity direction from the body's angular velocity and the points' relative position.
Apply the no-slip condition for a wheel rolling on a stationary surface.
Check velocity results using units, limiting cases, and geometric constraints.

1. The rigid-body velocity relation

A rigid body is an ideal body whose distances between all pairs of points remain constant. In planar motion, it may translate in the plane and rotate about an axis perpendicular to that plane. The observer in this lesson is fixed to the ground; the reference frame uses stationary

x

and

y

axes. A point's position is measured from the same frame, and its velocity is the time rate of change of that position.

Choose two points

A

and

B

on one body. The vector from

A

to

B

, written

\mathbf r_{B/A}

, points from

A

toward

B

. The relative-velocity equation says that

B

's velocity equals

A

's velocity plus the rotational contribution from the body's angular velocity. In a plane, the rotational contribution is perpendicular to

\mathbf r_{B/A}

, and its direction follows the right-hand rule.

Let

\omega

be the signed angular velocity: counterclockwise is positive and clockwise is negative. In component form, if

\mathbf r_{B/A}=x\mathbf i+y\mathbf j

, then the relative contribution has components

(-\omega y,\,\omega x)

. This is a useful sign check: for counterclockwise rotation, a point to the right of the reference point has an upward relative velocity.

\mathbf v_B=\mathbf v_A+\boldsymbol\omega\times\mathbf r_{B/A}

Use a single stationary observer and coordinate frame for all point velocities.
The position vector in the equation must point from the reference point to the point whose velocity is sought.
The rotational velocity is perpendicular to the line joining the two points.

2. Choosing points, signs, and useful constraints

The equation is vector-valued, so it supplies two scalar component equations in planar motion. Use known velocity information to choose the reference point. For example, if a point is fixed to a stationary pin, its velocity is zero; if a point slides along a straight guide, its velocity must lie along that guide. These facts reduce the unknowns.

For a planar body,

\boldsymbol\omega=\omega\mathbf k

, where

\mathbf k

points out of the page. The component form is

v_{Bx}=v_{Ax}-\omega y_{B/A}

and

v_{By}=v_{Ay}+\omega x_{B/A}

. The coordinate differences are signed: a point below

A

has negative

y_{B/A}

. Keeping these signs is safer than trying to guess the direction of each contribution.

If a body undergoes pure translation at an instant, its angular velocity is zero and every point has the same velocity. If it rotates about a fixed point, set that point's velocity to zero. The resulting relation says another point's velocity is tangent to its circular motion about the fixed point, with magnitude

|\omega|r

.

A useful special case is rolling without slipping on a stationary surface. At the instant of contact, the point on the wheel has zero velocity relative to the ground. This is a velocity constraint, not a statement that the wheel's centre is stationary. For a wheel of radius

R

moving right, counterclockwise is positive, so the contact-point condition gives

v_C+\omega R=0

; hence clockwise rotation accompanies rightward rolling.

v_{Bx}=v_{Ax}-\omega y_{B/A},\qquad v_{By}=v_{Ay}+\omega x_{B/A}

Apply guide, pin, or no-slip information as a velocity constraint.
Use signed coordinate differences and signed angular velocity consistently.
A fixed point has zero velocity; a rolling wheel's centre generally does not.

3. A reliable solution procedure

First identify the body and the observer. Sketch the body at the instant being analysed, mark the two points, and show the directed position vector from the chosen reference point. Add velocity arrows only where their directions or values are known. A rigid-body diagram is often more helpful here than a free-body diagram because this topic relates motion, not forces.

Next list what is known and unknown. Select positive axes and define positive rotation before substituting values. Write the vector equation, then resolve it into horizontal and vertical components. Use the geometry to convert any stated guide direction or rolling condition into component equations.

Solve symbolically before inserting numbers when practical. Give velocity units of metres per second and angular velocity units of radians per second. Since radians are dimensionless in this relation, multiplying angular velocity by a length gives velocity units. Finally check that the calculated direction agrees with the geometry and that special cases behave sensibly: zero rotation should make the point velocities equal, and zero reference-point velocity should leave only the rotational contribution.

\left[\omega r\right]=\mathrm{s^{-1}}\,\mathrm{m}=\mathrm{m/s}

Draw the relative-position vector in the direction required by the subscripts.
Resolve vector equations into components before solving for unknowns.
Check both units and the direction implied by the signs.

Worked example

A point on a rotating bar

A straight bar rotates counterclockwise at

4.0\ \mathrm{rad/s}

. At the instant shown, point

B

is

0.60\ \mathrm m

to the right and

0.20\ \mathrm m

above point

A

. Point

A

moves right at

1.5\ \mathrm{m/s}

. Find the velocity of

B

in the stationary ground frame.

Bar velocity relation

The sketch indicates the known velocity and the counterclockwise rotation; the point locations are stated in the problem.

Set the frame and geometry
Treat the bar as the system and measure velocities from the stationary ground frame. Take right and up as positive. The directed vector from $A$ to $B$ has components $0.60\ \mathrm m$ and $0.20\ \mathrm m$ .
$\mathbf r_{B/A}=(0.60\mathbf i+0.20\mathbf j)\ \mathrm m$
Write the relative-velocity components
The angular velocity is positive because the rotation is counterclockwise. The rotational contribution is therefore leftward from the point's upward offset and upward from its rightward offset.
$v_{Bx}=v_{Ax}-\omega y_{B/A},\quad v_{By}=v_{Ay}+\omega x_{B/A}$
Substitute and solve
Use the stated velocity of $A$ and the two position components. The result has a positive horizontal component and a positive vertical component.
$v_{Bx}=1.5-(4.0)(0.20)=0.70\ \mathrm{m/s},\quad v_{By}=0+(4.0)(0.60)=2.4\ \mathrm{m/s}$

Answer:

\mathbf v_B=(0.70\mathbf i+2.4\mathbf j)\ \mathrm{m/s}

. Its speed is approximately

2.50\ \mathrm{m/s}

, directed about

74^\circ

above the positive

x

-axis.

Check: Each rotational component has units

(\mathrm{rad/s})(\mathrm m)=\mathrm{m/s}

. The upward component is consistent with counterclockwise rotation at a point to the right of

A

.

Worked example

A slider connected to a rotating link

A link has point

A

fixed to ground and point

B

constrained to move horizontally. At one instant,

B

is

0.40\ \mathrm m

to the right and

0.30\ \mathrm m

above

A

. The link's angular velocity is

3.0\ \mathrm{rad/s}

counterclockwise. Determine the horizontal velocity of

B

and the vertical velocity of

A

.

Link with fixed pivot and slider

Point A is fixed; the slider at B can move only horizontally.

Identify point constraints
Use the ground frame with positive $x$ right and positive $y$ up. Since $A$ is fixed, both components of its velocity are zero. Since $B$ slides horizontally, its vertical velocity is zero. The requested horizontal velocity can be found from the vertical component of the rigid-body relation.
$\mathbf v_A=\mathbf 0,\qquad v_{By}=0$
Use the geometry and rotation sign
The vector from $A$ to $B$ has $x_{B/A}=0.40\ \mathrm m$ and $y_{B/A}=0.30\ \mathrm m$ . Counterclockwise angular velocity is positive.
$v_{By}=v_{Ay}+\omega x_{B/A}$
Apply the constraint
The vertical component equation confirms the slider constraint: it gives zero vertical velocity. The horizontal component is found from the other component equation.
$0=0+(3.0)(0.40),\quad v_{Bx}=0-(3.0)(0.30)=-0.90\ \mathrm{m/s}$

Answer: Point

B

moves left at

0.90\ \mathrm{m/s}

, and point

A

has zero vertical velocity because it is fixed.

Check: The calculated horizontal direction is left: the point

B

is above the fixed pivot while the link rotates counterclockwise. The link length does not change; the velocity of

B

is perpendicular to

AB

.

Worked example

Rolling wheel and the ground-contact point

A wheel of radius

0.25\ \mathrm m

rolls to the right without slipping on a stationary horizontal floor. Its centre

C

moves right at

2.0\ \mathrm{m/s}

. Find the wheel's angular velocity and the ground-frame velocity of its top point

T

.

Wheel velocity points

The no-slip condition applies at the instantaneous contact point with the stationary floor.

Set the rolling condition
Use the ground frame with positive $x$ to the right and counterclockwise angular velocity positive. The contact point $P$ has zero velocity relative to the stationary floor. Relative to the centre, it is one radius below, so its rotational velocity is horizontal.
$0=v_{Px}=v_C+\omega R$
Find the angular velocity
Solving the no-slip equation gives a negative angular velocity, which means clockwise rotation. The negative sign is required for a wheel rolling right.
$\omega=-\frac{v_C}{R}=-\frac{2.0}{0.25}=-8.0\ \mathrm{rad/s}$
Find the top-point velocity
The top point is one radius above the centre. Its rotational velocity relative to the centre is to the right for clockwise rotation. Add that contribution to the centre's rightward velocity.
$v_{Tx}=v_C-\omega R=2.0-(-8.0)(0.25)=4.0\ \mathrm{m/s}$

Answer: The wheel rotates clockwise at

8.0\ \mathrm{rad/s}

. The top point moves right at

4.0\ \mathrm{m/s}

relative to the ground.

Check: The contact point has zero velocity because

2.0+(-8.0)(0.25)=0

. The top point moves twice as fast as the centre in this ideal rolling case.

Common mistakes and how to avoid them

Using the position vector from

B

to

A

while applying the equation for

\mathbf v_B

relative to

A

.

Correction: Write the subscripts first:

\mathbf r_{B/A}

points from

A

to

B

.

Treating counterclockwise rotation as negative or changing the sign convention midway.

Correction: State the convention at the start and keep counterclockwise positive throughout.

Assuming every point on a rigid body has the same velocity.

Correction: Only pure translation makes all point velocities equal; rotation adds a point-dependent contribution.

Setting the wheel centre's velocity to zero because the wheel touches the ground.

Correction: For rolling without slipping, the contact point has zero ground-frame velocity; the centre moves with the wheel.

Lesson summary

For two points on one planar rigid body, add the rotational velocity contribution to the reference point's velocity.
Use a directed position vector from the reference point to the point being analysed.
In components, the rotational contribution is $(-\omega y,\,\omega x)$ .
Apply fixed-point, guide, and no-slip conditions to determine unknown velocities.
Check signs, directions, units, and the zero-rotation or fixed-point limits.

Check your understanding

Question 1

A body rotates counterclockwise. Point

B

is directly above point

A

. What is the direction of

B

's velocity relative to

A

?

Right
Left
Up
Down

Show answer and explanation

Right

For a point directly above

A

, the relative position has positive

y

and zero

x

. The component relation gives a negative horizontal component, wait: counterclockwise rotation at the top of a circle points left. Therefore the correct direction is left.

Question 2

A rigid body undergoes pure translation at an instant. What is the relationship between the velocities of any two of its points?

They are equal as vectors.
They have equal speed but opposite directions.
They are perpendicular.
One must be zero.

Show answer and explanation

They are equal as vectors.

For pure translation, angular velocity is zero. The relative-velocity equation then gives identical velocity vectors for all points.

Question 3

A wheel of radius

0.20\ \mathrm m

rolls right without slipping, with its centre moving at

1.6\ \mathrm{m/s}

. What is its signed angular velocity when counterclockwise is positive?

$+8.0\ \mathrm{rad/s}$
$-8.0\ \mathrm{rad/s}$
$+0.125\ \mathrm{rad/s}$
$-0.125\ \mathrm{rad/s}$

Show answer and explanation

-8.0\ \mathrm{rad/s}

No slip requires

0=v_C+\omega R

, so

\omega=-v_C/R=-1.6/0.20=-8.0\ \mathrm{rad/s}

. The negative sign denotes clockwise rotation.

Key terms

Rigid body: An ideal body whose distances between its points remain constant.
Relative position vector: The directed vector from a chosen reference point to another point on the same body.
Angular velocity: The signed rate of rotation; in this lesson counterclockwise is positive.
No-slip rolling: Rolling on a stationary surface with zero ground-frame velocity at the instantaneous contact point.

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9.3 · Relate velocities of points on a planar rigid body

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 9.3

What you will learn

1. The rigid-body velocity relation

2. Choosing points, signs, and useful constraints

3. A reliable solution procedure

A point on a rotating bar

A slider connected to a rotating link

Rolling wheel and the ground-contact point

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review