EQUATIONS OF MOTION: GENERAL PLANE MOTIONfacstaff.cbu.edu/pshiue/Courses/ME202/Notes/Ch17_5.pdfEQUATIONS OF MOTION: GENERAL PLANE MOTION Today’s Objectives: Students will be able

EQUATIONS OF MOTION:

GENERAL PLANE MOTION Today’s Objectives:

Students will be able to:

1. Analyze the planar kinetics of a

rigid body undergoing general

plane motion. In-Class Activities:

• Check Homework

• Reading Quiz

• Applications

• Equations of Motion

• Frictional Rolling Problems

• Concept Quiz

• Group Problem Solving

• Attention Quiz

READING QUIZ

1. If a disk rolls on a rough surface without slipping, the

acceleration af the center of gravity (G) will _________ and

the friction force will be __________.

A) not be equal to a r; less than sN

B) be equal to a r; equal to kN

C) be equal to a r; less than sN

D) None of the above

2. If a rigid body experiences general plane motion, the sum of

the moments of external forces acting on the body about any

point P is equal to __________.

A) IP a B) IP a + maP

C) m aG D) IG a + rGP × maP

APPLICATIONS

As the soil compactor accelerates

forward, the front roller experiences

general plane motion (both translation

and rotation).

The forces shown on the

roller’s FBD cause the

accelerations shown on the

kinetic diagram.

Is the point A the IC?

=

What are the loads experienced by

the roller shaft or bearings?

APPLICATIONS (continued)

The lawn roller is pushed forward with a force of 200 N when

the handle is at 45°.

How can we determine its translation acceleration and angular

acceleration?

Does the acceleration depend on the coefficient’s of static and

kinetic friction?

APPLICATIONS (continued)

During an impact, the center of

gravity of this crash dummy will

decelerate with the vehicle, but also

experience another acceleration due

to its rotation about point A.

Why?

How can engineers use this information to determine the forces

exerted by the seat belt on a passenger during a crash?

EQUATIONS OF MOTION: GENERAL PLANE MOTION

(Section 17.5) When a rigid body is subjected to external

forces and couple-moments, it can undergo

both translational motion and rotational

motion. This combination is called general

plane motion.

Fx = m (aG)x

Fy = m (aG)y

MG = IG a

Using an x-y inertial coordinate system,

the equations of motions about the

center of mass, G, may be written as:

Sometimes, it may be convenient to write the

moment equation about a point P other than G.

Then the equations of motion are written as

follows:

Fx = m (aG)x

Fy = m (aG)y

MP = (Mk )P

In this case, (Mk )P represents the sum of the

moments of IGa and maG about point P.

EQUATIONS OF MOTION: GENERAL PLANE MOTION

(continued)

FRICTIONAL ROLLING PROBLEMS

When analyzing the rolling motion of wheels, cylinders, or disks,

it may not be known if the body rolls without slipping or if it

slides as it rolls.

The equations of motion will be:

Fx = m(aG)x => P - F = m aG

Fy = m(aG)y => N - mg = 0

MG = IGa => F r = IG a

There are 4 unknowns (F, N, a, and aG) in

these three equations.

For example, consider a disk with mass m

and radius r, subjected to a known force P.

Hence, we have to make an assumption

to provide another equation. Then, we

can solve for the unknowns.

The 4th equation can be obtained from

the slip or non-slip condition of the disk.

Case 1:

Assume no slipping and use aG = a r as the 4th equation and

DO NOT use Ff = sN. After solving, you will need to verify

that the assumption was correct by checking if Ff sN.

Case 2:

Assume slipping and use Ff = kN as the 4th equation.

In this case, aG ar.

FRICTIONAL ROLLING PROBLEMS

(continued)

PROCEDURE FOR ANALYSIS

Problems involving the kinetics of a rigid body undergoing

general plane motion can be solved using the following procedure.

1. Establish the x-y inertial coordinate system. Draw both the

free body diagram and kinetic diagram for the body.

2. Specify the direction and sense of the acceleration of the

mass center, aG, and the angular acceleration a of the body.

If necessary, compute the body’s mass moment of inertia IG.

3. If the moment equation Mp= (Mk)p is used, use the

kinetic diagram to help visualize the moments developed by

the components m(aG)x, m(aG)y, and IGa.

4. Apply the three equations of motion.

PROCEDURE FOR ANALYSIS

(continued)

6. Use kinematic equations as necessary to complete the

solution.

5. Identify the unknowns. If necessary (i.e., there are four

unknowns), make your slip-no slip assumption (typically no

slipping, or the use of aG = a r, is assumed first).

Key points to consider:

1. Be consistent in using the assumed directions.

The direction of aG must be consistent with a.

2. If Ff = kN is used, Ff must oppose the motion. As a test,

assume no friction and observe the resulting motion.

This may help visualize the correct direction of Ff.

7. If a slip-no slip assumption was made, check its validity!!!

Find: The angular acceleration (a) of the spool and the tension

in the cable.

Plan: Focus on the spool. Follow the solution procedure (draw

a FBD, etc.) and identify the unknowns.

EXAMPLE

Given: A spool has a mass of 200 kg and a radius of gyration (kG)

of 0.3 m. The coefficient of kinetic friction between the

spool and the ground is k = 0.1.

EXAMPLE (continued) Solution:

The free body diagram and kinetic diagram for the body are:

Equations of motion:

Fy = m (aG)y : NB − 1962 = 0

NB = 1962 N

IG a

maG

=

1962 N

EXAMPLE

(continued)

Fx = m (aG)x: T – 0.1 NB = 200 aG = 200 (0.4) a

T – 196.2 = 80 a

MG = IG a : 450 – T(0.4) – 0.1 NB (0.6) = 20 (0.3)2 a

450 – T(0.4) – 196.2 (0.6) = 1.8 a

Solving these two equations, we get

a = 7.50 rad/s2, T = 797 N

Note that aG = (0.4) a. Why ?

CONCEPT QUIZ

2. For the situation above, the moment equation about G is?

A) 0.75 (FfA) - 0.2(30) = - (80)(0.32)a

B) -0.2(30) = - (80)(0.32)a

C) 0.75 (FfA) - 0.2(30) = - (80)(0.32)a + 80aG

D) None of the above

1. An 80 kg spool (kG = 0.3 m) is on a

rough surface and a cable exerts a 30 N

load to the right. The friction force at A

acts to the __________ and the aG

should be directed to the __________ .

A) right, left B) left, right

C) right, right D) left, left

30N

A

0.2m

0.75m

a

G •

Find: The angular acceleration if

s = 0.12 and k = 0.1.

Plan: Follow the problem solving

procedure.

Given: A 80 kg lawn roller has a

radius of gyration of

kG = 0.175 m. It is pushed

forward with a force of 200 N.

GROUP PROBLEM SOLVING

Solution:

The moment of inertia of the roller about G is

IG = m(kG)2 = (80)(0.175)2 = 2.45 kg·m2

Equations of motion:

Fx = m(aG)x

FA – 200 cos 45 = 80 aG

Fy = m(aG)y

NA – 784.8 – 200 sin 45 = 0

MG = IG a

– 0.2 FA = 2.45 a

Another equation is needed to allow solving for the

unknowns.

GROUP PROBLEM SOLVING (continued)

FBD:

We have 4 unknowns: NA, FA, aG and a.

The three equations we have now are:

FA – 200 cos 45° = 80 aG

NA – 784.8 – 200 sin 45° = 0

– 0.2 FA= 2.45 a

Now solving the four equations yields:

NA = 926.2 N, FA = 61.4 N, a = -5.01 rad/s2 , aG = -1.0 m/s2

The no-slip assumption must be checked.

Is FA = 61.4 N s NA= 111.1 N ?

Yes, therefore, the wheel rolls without slip.

First, assume the wheel is not slipping.

Thus, we can write

aG = r a = 0.2 a

GROUP PROBLEM SOLVING (continued)

ATTENTION QUIZ

2. Select the equation that best represents the “no-slip”

assumption.

A) Ff = s N B) Ff = k N

C) aG = r a D) None of the above

1. A slender 100 kg beam is suspended by a

cable. The moment equation about point A is?

A) 3(10) = 1/12(100)(42) a

B) 3(10) = 1/3(100)(42) a

C) 3(10) = 1/12(100)(42) a + (100 aGx)(2)

D) None of the above

4m

10 N

3m

A