Chapter 10 Class 11

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Chapter 10

Instructor : Dr. Iyad Saadeddin

Chapter 10: Rotation of Rigid Objects About a Fixed Axis       

Angular position, velocity, acceleration Constant angular acceleration Rotational kinetic energy Moments of inertia Torque Work, power and energy Rolling motion

Rotational Motion Particle on rigid object at point p rotates throug an angle θ

Arc length s:



s r

s  r

in radians (rad.)

Full circle = 360° = 2 rad. 2 rad ≡ 1 revolution (‫)دورة‬ Degree 180 → θ →

rad. π ??

 (rad) 

 180

 (deg)

π = 22/7 = 3.14

Angular Quantities Angular displacement: Average angular speed: Instantaneous angular speed:

   f   i 

 f  i

  lim t  0

t f  ti



 (Rad/s)= s -1 t

 d  t dt

  i  Average angular  f  acceleration: t f  ti t Instantaneous angular accel eration:

  lim t  0

(Rad.)

Angular Velocity As particle moves from angular position ?i to ?f



 (Rad/s)= s -1

(Rad/s²)= s -2

  d  (Rad/s²)= s -2 t dt

All particles of a rigid object rotate at the same angular displacement, speed and acceleration.



Angular velocity , is a vector. For rotation about a fixed axis, the direction of the velocity is along the axis of rotation. Use the right hand rule to determine direction.

For rotation counterclockwise (‫)ﺑﻌﻛس ﻋﻘﺎرب اﻟﺳﺎﻋﺔ‬  +ve ω

For rotation clockwise ( ‫ﻣﻊ ﻋﻘﺎرب‬ ‫ )اﻟﺳﺎﻋﺔ‬ -ve ω


Rotational Kinematics For rotational motion with constant rotational acceleration  The equations of motion are similar in to the equation of motion in one dimension (1D); Only do the following symbol replacement θ2 ω= 7200 rev/min

90° R=6 cm

x≡θ

θ1

v≡ω a≡α

Rotational Kinematics

Example:

Linear (1D) Motion

Rotational Motion

with constant linear acceleration, a

with constant rotational acceleration, 

v f  vi  at

 f   i  t

x f  xi  12 (vi  v f )t

 f   i  12 ( i   f )t

x f  xi  x  12 (vi  v f )t

 f   i    12 (i   f )t

1 2

x f  xi  x  (vi  v f )t

1 x  vi t  a t 2 2 2

2

α

v f  vi  2a( x f  xi )

  3.50 rad s 2

1   i t  t 2 2 2

2

 f   i  2 ( f   i )

  3.5rad / s 2 , ti  0s, t  2.00 rad s , t f  2.00 s i

  ?, revolution s  ?, t f  ?


Example:

Angular and Linear Quantities

  3.5rad / s 2 , ti  0 s, t  2.00 rad s , t f  2.00s i

  ?, revolutions  ?, t f  ?

 1 rev.    11 rad    1.75 rev.  2 rad . 

Arc length s:

s  r

Tangential speed of a point P:

v  r

Tangential acceleration of a point P:

at  r

Centripetal acceleration for rotation object:

v2 ar   r 2 r 2

a  ar  at2

Example: 

Example:

A race car accelerates uniformly from a speed of 40 m/s to 60 m/s in 5 s around a circular track of radius 400 m. When the car reaches a speed of 50 m/s find the  Centripetal acceleration,  Angular speed,  Tangential acceleration,  And the magnitude of the total acceleration.

At v = 50 m/s find ac=?, ω=?,

at =?,

a

at =?,

v 2 (50) 2 ac    6.25 m/s2 r 400 v 50 v  r      0.125 rad/s r 400 v2  v1  at t  at 

v1=40 m/s, v2=60m/s, and t=5s At v = 50 m/s find ac=?, ω=?,

v1=40 m/s, v2=60m/s, and t=5s

v2  v1 60  40   4 m/s t 5

a  ac2  at2  6.252  4 2  7.42 m/s 2

a


Rotational Energy

Example:

A rigid object made up of a collection of i particles with mass mi has a rotational kinetic energy of:

K

1 1 mi vi2   mi ri 2 2  2 2

1 2 I 2

KR 

I≡m

(J)

ω≡v

where I is the moment of inertia:

I   mi ri i

2

Spheres of mass m has I=0 because r=0, they lie on y-axis

(kg.m2) (for collection of particles) r is the distance from rotational axis

Example:

Calculating Moments of Inertia for collection of particles:

I   mi ri

2

i

For an extended, rigid object:

I   r 2 dm

m=ρV  dm= ρdV m=σA  dm=σdA m=λL  dm=λdL

dm  dV

I   r 2 dV


Moments of Inertia of Various Objects

Example:

Extended object 

ICM : Moment of inertia about an axis of rotation through the center of mass

; λdL=λdx because L is on x-axies

Parallel Axis Theorem

Torque Fsin

If ICM is known, the moment of inertia through a parallel axis of rotation a distance D away from the center of mass is:

I  I CM  MD 2

F

r



Fcos

I  I CM  MD 2 2

I

 Is angle between F and r directions

1  L 1 ML2  M    ML2 12 3 2   

Consider a rigid object about a pivot point (‫)ﻧﻘطﺔ ارﺗﻛﺎز‬. A force is applied to the object. This force causes the object to rotate having what is called Torque τ.

  rF sin 


Torque Torqueand Angular Acceleration

Example: Two forces T1 and T2 are applied as shown

Ft  mat

at  r   Ft r

For rotation counterclockwise (‫)ﺑﻌﻛس ﻋﻘﺎرب اﻟﺳﺎﻋﺔ‬  +ve α

I  mr 2

  I

  I

For rotation clockwise ( ‫ﻣﻊ ﻋﻘﺎرب‬ ‫ )اﻟﺳﺎﻋﺔ‬ -ve α

Example:A uniform rod of length L and mass M is attached as shown. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end? a=? and at=? * The rode will move like pendulum under the effect of F g=Mg

α=? T=?

a=?

?Extended object  look at the CM solution

T=? a

L   rF sin   rF  ( ) Mg 2

L but   I  ( ) Mg 2  ( L / 2) Mg 3g     2 I 1 / 3ML 2L

at

The translational acceleration is


Work, Power and Energy Work in linear motion

dW  F  ds W  F  s  Fs cos 

P 

dW  Fv dt

Work in rotational motion

dW  F  ds dW  τdθ W   P  


Work - Kinetic Energy Theorem The work-kinetic energy theorem for linear motion:

1 2

1 2

2 2  W  mv f  mvi

External work done on an object changes its kinetic energy

… and for rotational motion:

1 2

Rolling Motion Pure rolling motion: object rotating at angular speed ω with No slipping occurs

1 2

2 2  W  I f  I i

External rotational work done on an object changes its rotational kinetic energy

Rolling Motion

Rolling Motion

The linear speed of the center of mass

vCM 

ds d R  R dt dt

+

=

The linear acceleration of the center of mass

aCM

Pure rolling motion is a superposition of pure translation and pure rotation motions

dv d  CM  R  R dt dt

While the angular velocity (and acceleration) of any point of the object is the same, the linear speed changes.

vP  0 vP  2vCM

Kinetic energy of rolling motion

K=KT + KR 1 1 K  Mv 2  I CM  2 2 2


Problem

Review 





KR+KT



Linear quantities have analogous angular counterparts. Torque is the tendency of a force to rotate an object. The total kinetic energy of a rotating object has to include its rotational kinetic energy. Pure rolling motion is the superposition of pure translational and pure rotational motion.

Chapter 10 Class 11

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