DYNAMICS OF A RIGID BODY

General Definition of Torque• Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by

• = 0° or = 180 °: torque are equal to zero• = 90° or = 270 °: magnitude of torque attain to the maximum

sinrF=

Torque Units and Direction• The SI units of torque are N.m• Torque is a vector quantity• Torque magnitude is given by

• Torque will have direction• If the turning tendency of the force is counterclockwise, the

torque will be positive• If the turning tendency is clockwise, the torque will be

negative

The work done by the torque is given by

sinrF=

ddW =

Net Torque• The force will tend to cause a counterclockwise rotation about O

• The force will tend to cause a clockwise rotation about O

• S = 1 + 2 = F1d1 – F2d2

• If S 0, starts rotating • If S = 0, rotation rate does not change

2F

1F

Rate of rotation of an object does not change, unless the object is acted on by a net torque

Power delivered by torque• To find the instantaneous power delivered by the torque, divide both sides by dt

or

ddW =

dtd

dtdWP ==

=P

Newton’s Second Law for a Rotating Object

• When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration

• The angular acceleration is directly proportional to the net torque

• The angular acceleration is inversely proportional to the moment of inertia of the object

• The relationship is analogous to =maF

I S =

Extended Work-Energy Theorem • The work-energy theorem tells us

• When Wnc = 0,

• The total mechanical energy is conserved and remains the same at all times

• Remember, this is for conservative forces, no dissipative forces such as friction can be present

ffii mgymvmgymv +=+ 22

21

21

iiff KEPEPEKE +=+

PEKEWnc +=

Total Energy of a System• A ball is rolling down a ramp• Described by three types of energy

• Gravitational potential energy

• Translational kinetic energy

• Rotational kinetic energy

• Total energy of a system22

21

21 IMghMvE ++=

2

21 IKEr =

MghPE =

2

21 MvKEt =

Conservation of Mechanical Energy• Conservation of Mechanical Energy

• Remember, this is for conservative forces, no dissipative forces such as friction can be present

2222

21

21

21

21

fffiii ImgymvImgymv ++=++

( ) ( )t r g i t r g fKE KE PE KE KE PE+ + = + +

Work-Energy in a Rotating System• The work done on the body by the external torque equals the change in the rotational kinetic energy

• The work equals the negative of the change in potential energy

• Conservation of Energy in Rotational Motion

2IKW ==

21

2212 2

121 IIKKW ==

21

2212 2

121 IIUU =+

constant21 2 =+= IUE

Problem Solving Hints• Choose two points of interest

• One where all the necessary information is given• The other where information is desired

• Identify the conservative and non-conservative forces

• Write the general equation for the Work-Energy theorem if there are non-conservative forces• Use Conservation of Energy if there are no non-

conservative forces• Use v = to combine terms• Solve for the unknown

Example• A meterstick is initially standing vertically on the floor. If the falls over, with what angular velocity will it hi the floor? Moment of inertia is Ml2/2l = 1.0 mU = mgy y0 = com = .5 m

0 = 0y0 = com = 0 m

Therefore,

What rate is gravity delivering energy? The mass of the meterstick is 0.15kgR= l/2

6/21 222

2 MlI =

2/061

02

02 MglMgyMlE +=+=

22

22

22

610

61 MlMgyMlE +=+=

2/61 2

22 MglMl = lg /32

2 = radlg 4.50.1/)81.9(3/32 ===

=P

mNmgRF ==== 74.0)1)(5.0)(81.9)(15.0(90sin21sin

WP 0.4)4.5)(74.0( ==

General Problem Solving Hints• The same basic techniques that were used in linear motion can be applied to rotational motion.

• F becomes • m becomes I • a becomes • v becomes ω • x becomes θ

Angular Momentum• Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum

• Linear momentum is defined as

• Angular momentum is defined as

IL =

mvp =

Lt

S =

Angular Momentum and Torque• Net torque acting on an object is equal to the time rate of change of the object’s angular momentum

• Angular momentum is defined as

• Analog in impulse

tpF

=

tII

tI

tII

=

=

== 00 )(

Lt

S =

tL

==interval time

momentumangular in change

Angular Momentum Conservation• If the net torque is zero, the angular momentum remains constant

• Conservation of Angular Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero.• That is, when

• then

0=

=tL

0=L0, i fi i ffL L or I I S = = =

fi LL = ffii II =

Blocks and Pulley• Two blocks with masses m1 = 5 kg and m2

= 7 kg are attached by a string over a pulley with mass M = 2kg. The pulley, which runs on a frictionless axle, is a hollow cylinder with radius 0.05 m over which the string moves without slipping.

Blocks and Pulley• Two blocks with masses m1 = 5 kg and m2

= 7 kg are attached by a string over a pulley with mass M = 2kg. The pulley, which runs on a frictionless axle, is a hollow cylinder with radius 0.05 m over which the string moves without slipping.

gmTam 111 = gmTam 222 =

21 RTRTI net +== Ra =

)()()/( 2211 amgmRamgmRRaI ++=

gRImm

mmRa)/()(

)/ 221

22

++

==

• An automobile with rear-wheel drive is accelerating at 4.0m/s2 along a straight road. Consider one of the wheels of this automobile. The axle pushes forward, providing an acceleration of 4.0 m/s2. Simultaneously, the friction force of the road pushes the bottom of the wheel backward, providing a torque that gives the wheel an angular acceleration. The wheel has a radius of 0.38m and a mass of 25kg. Find the backward force that the friction force exerts on the wheel, and find the forward force that the axle exerts on the wheel.

RfI = fMR =21

Ra = fMa =21

Nsmkgf 50)/0.4)(25(21 2 ==