Chapter9 Rotational Dynamics Translational and Rotational Motion Torque

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Chapter-9 Rotational Dynamics

Chapter-9 Rotational Dynamics

Translational and Rotational Motion

Translational and Rotational Motion

Torque Which one of the above is the easiest to open a door?

Torque Which one of the above is the easiest to open a door?

Definition of Torque is a vector quantity. Direction: The torque is positive when the

Definition of Torque is a vector quantity. Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation. SI Unit of Torque: newton · meter (N · m)

Example 1 Different Lever Arms, Different Torques In Figure 9. 3 a force whose

Example 1 Different Lever Arms, Different Torques In Figure 9. 3 a force whose magnitude is 55 N is applied to a door. However, the lever arms are different in the three parts of the drawing. Find the magnitude of the torque in each case.

The Achilles Tendon

The Achilles Tendon

9. 2 Rigid Objects in Equilibrium

9. 2 Rigid Objects in Equilibrium

Equilibrium Of A Rigid Body A rigid body is in equilibrium if it has

Equilibrium Of A Rigid Body A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero:

Applying the Conditions of Equilibrium to a Rigid Body 1. Select the object to

Applying the Conditions of Equilibrium to a Rigid Body 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all the external forces acting on the object. 3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4. Apply the equations that specify the balance of forces at equilibrium: SFx = 0 and SFy = 0. 5. Select a convenient axis of rotation. Identify the point where each external force acts on the object, and calculate the torque produced by each force about the axis of rotation. Set the sum of the torques about this axis equal to zero: St = 0. 6. Solve the equations for the desired unknown quantities.

Example 4 Fighting a Fire In Figure 9. 7 a an 8. 00 -m

Example 4 Fighting a Fire In Figure 9. 7 a an 8. 00 -m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N, stands 6. 30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

9. 3. Center of Gravity The center of gravity of a rigid body is

9. 3. Center of Gravity The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated.

Uniform Thin Rod The center of gravity of the rod is at its geometrical

Uniform Thin Rod The center of gravity of the rod is at its geometrical center.

Center of Gravity of a Group of Objects

Center of Gravity of a Group of Objects

Example 6 The Center of Gravity of an Arm The horizontal arm in Figure

Example 6 The Center of Gravity of an Arm The horizontal arm in Figure 9. 11 is composed of three parts: the upper arm (weight W 1 = 17 N), the lower arm (W 2 = 11 N), and the hand (W 3 = 4. 2 N). The drawing shows the center of gravity of each part, measured with respect to the shoulder joint. Find the center of gravity of the entire arm, relative to the shoulder joint.

Overloading a Cargo Plane (b) Correctly loaded (c) Incorrectly loaded

Overloading a Cargo Plane (b) Correctly loaded (c) Incorrectly loaded