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Planar Kinematics of Rigid Bodies When a rigid body is rotating around an axis, the rigid body’s angular velocity, or rate of rotation, and its angular acceleration are

� � ����

� � ���� � ���

���

where is the angle between the reference line and the body-fixed line described the position, or orientation, of the rigid about the rotation axis. To describe the orientation of the rotation axis, we need to introduce the so-called angular velocity vector, denoted by , with magnitude given in the equation above. The vector is parallel to the axis of rotation. Furthermore, the direction of is related to the direction of the body’s rotation through the right-hand rule: Pointing the thumb of the right hand in the direction of , the fingers curl around in the direction of rotation (refer to the figure below) If the movement of the body is confined within the x-y plane, then we have � �� � � �� Hence, by definition, � is positive when the rotation is anti-clockwise. General Motions: Velocities Let’s start by considering the velocities of two points A and B on a rigid body moving in the x-y plane. The position vector of A can be expressed as � � � � �/� with �/� being the position vector of A relative to B. Taking the derivative w.r.t. time leads to �� � �� � ��/� where ��/� � � �/�/�� is the velocity of A relative to B.

Furthermore, if the angular velocity vector of the rigid body is

��/� � � �/��� � So, we have �� � �� � � Example: Rolling Disk. Consider a circular disk of radius angular velocity � on the x-y plane. By rolling, we mean that the velocity of the disk relative to the surface is zero at the point of contact That is �� � �, the velocity of point center of the disk of then be found as �� � �� � � �/� � ��

� � � � � � �0 0 �0 � 0� � � Similarly, the velocity of point In the Figure, is �� � �� � � �/�

� ���� � � � � �0 0 �� 0 0� � Example: Suppose that bar ABangular velocity of 10 rad/s. Determine the angular velocity of bar C.

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Furthermore, if the angular velocity vector of the rigid body is , then we have

� � �/�.

� �/�.

. Consider a circular disk of radius R rolling with counterplane. By rolling, we mean that the velocity of the disk relative to

the surface is zero at the point of contact C.

the velocity of point B, the center of the disk of then be found as

� ��! � ��!

����. Similarly, the velocity of point A, as shown

� � ���� � ���.

AB in the linkage system shown below rotates with a clockwise angular velocity of 10 rad/s. Determine the angular velocity of bar BC and the velocity of point

, then we have

rolling with counter-clockwise plane. By rolling, we mean that the velocity of the disk relative to

in the linkage system shown below rotates with a clockwise and the velocity of point

Solutions: We have �/� � 0.4� � 0.4� �� � �� � �� �

� 0 � � � �0 00.4 0.4 In addition, we also have �/� �� � �� � �� �

⇒ &�� � 4� � '� � �0 � 4 � 0.4��� from which we find �� � 5� and � Example: Bar AB in the mechanicalvelocity of 10 rad/s. What is the vertical velocity

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(m), �� � �10� (rad/s)

�/�

��104 0 � � 4� � '� (m/s).

� � 0.8� � 0.4� (m), �� � ����, and �� � �/�

� � � �0 0 ���0.8 �0.4 0 � ��!� � 4 � 0.8���!�,

�� � 6�

mechanical system shown below rotates with a clockwise angular s. What is the vertical velocity &, of the rack-and-pinion gear?

� &��, hence

system shown below rotates with a clockwise angular pinion gear?

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Solutions: Notice that �/� = 0.5� + � (ft), �� = −10� (rad/s) �� = �� + �� × �/�

= 0 + � � � �0 0 −100.5 1 0 � = 10� − 5� (ft/s).

In addition, use �/- = −0.5� + 0.833� (ft), and let �- = −��-�, we have �� = �- + �- × �/-

= 0 + � � � �0 0 −��-−0.5 0.833 0 � = 0.833��-� + 0.5��-�. Next, recall that �/� = 1.333� − 0.167� (ft), and let �� = ����, we have �� = �� + �� × �/�

= �� + � � � �0 0 ���1.333 −0.167 0 � = �� + 0.167���� + 1.333����. Substituting the expressions of �� and �� into the above equation, we find 0.833��-� + 0.5��-� = 10 + 0.167���!� + −5 + 1.333���!� ⇒ ���=8.92 rad/s ��-=13.78 rad/s. Finally, the vertical velocity of the rack is equal to the velocity of the gear where it contacts the rack: &, = 0.5 ft!��- = 6.89 ft/s.

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General Motions: Accelerations Now, let’s consider the relationship between the accelerations of two points A and B on a rigid body. Recall that �� = �� + ��/� � �� � � �/�. Taking derivative w.r.t. time on both sides leads to

8� � 8� � 8�/� � 9���� � ���/��� .

Notice that

8�/� � ���/��� � ��� � �/� � � ��/�

� � � �/� � � : � �/�;, hence, we have 8� � 8� � � � �/� � � : � �/�;. For planar motion (i.e. � �� and � � ��), the above equation reduces to 8� � 8� � � � �/� � �< �/�. Example: The rolling disk, with radius R, in Figure below has counter-clockwise angular velocity � and counter-clockwise angular acceleration �. Determine the acceleration of point A.

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Solutions: The acceleration at the center B is 8� = −���, 8� = 8� + � × �/� − �< �/�. Using �/� = �� and � = ��, we have

8� = −��� + � � � �0 0 �� 0 0� − �<�� = −�� − �<�!� + ��� Example: Suppose that bar AB in the linkage system shown below has a counterclockwise angular velocity of 10 rad/s and a clockwise angular acceleration of 300 rad/s2. What are the angular accelerations of bars BC and CD? Solutions: The velocity of B is �� = �� + �� × �/� = � + 10�! × 2�! = −20� m/s!.

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Let ��-be the unknown angular velocity of bar CD. The velocity of C is �� = �- + �- × �/-

= � + � � � �0 0 ��-−2 2 0 � = −2��-� − 2��-�. On the other hand, we can also obtain �� in terms of The velocity of B: �� = �� + �� × �/� = −20� + ����! × 2�! = −20� + 2����, with ��� being the unknown angular velocity of bar BC. Equating two expressions of ��, we have ��- = 10 rad/s and ��� = −10 rad/s . We can use the same sequence of steps to determine the angular accelerations. The acceleration of B is 8� = 8� + ��� × �/� − ���< �/� = � + −300�! × 2�! − 10!< 2�! = 600� − 200� m/s!. The acceleration of C in terms of the acceleration of D is

8� = 8- + ��- × �/- − ��-< �/- = � + � � � �0 0 ��-−2 2 0 � − 10!< −2� + 2�!

= 200 − 2��-!� − 200 + 2��-!�. The acceleration of C in terms of the acceleration of B is

8� = 8� + ��� × �/� − ���< �/� = 600� − 200� + �� � �0 0 ���2 0 0 � − 10!< 2�!

= 400� − 200 − 2���!�. Equating two expressions leads to ��- = −100 rad/s< and ��� = 100 rad/s<.

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Practice Problems 1. The diameter of the disk is 1 m, and the length of bar AB is 1 m as shown in the figure below. The disk is rolling, and point B slides on the ground. Determine the angular velocity of bar AB and the velocity of point B. (Answers: ��� = 2.31 rad/s clockwise, &� = 3.15 m/s to the left) 2. Horizontal motion of the piston rod of the hydraulic cylinder controls the rotation of link OB about O. For the instant represented, &� = 2 m/s and OB is horizontal. Determine the angular velocity � of OB for this instant. (Answers: � = 8.59 rad/s counter-clockwise) 3. A linkage system is shown below. For the instant when = ? = 60°, the hydraulic cylinder gives A a velocity &� = 4 ft/s which is increasing by 3 ft/s each second. For this instant determine the angular acceleration of link BC. (Answers: ��� = 100 rad/s< counter-clockwise)

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4. For the linkage system shown below, if OA has a constant counter-clockwise angular velocity �A = 10 rad/s, calculate the angular velocities of AB and BC, as well as the angular acceleration of AB, for the position where the coordinates of A are B = −60 mm and C = 80 mm. Link BC is vertical for this position. (Answers: �� = 2.5� rad/s, �� = 5.83� rad/s, and ��� = 10.42� rad/s<) 5. The elements of a power hacksaw are shown in the figure below. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counter-clockwise speed of 60 rev/min, determine the acceleration of the blade for the position where = 90°, and find the corresponding angular acceleration of the link AB. (Answers: �� = 4.89 m/s<, and ��� = 0.467 rad/s< counter-clockwise)