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Torque, Equilibrium, and Stability Dual Credit Physics Montwood High School R. Casao.

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Presentation on theme: "Torque, Equilibrium, and Stability Dual Credit Physics Montwood High School R. Casao."— Presentation transcript:

1 Torque, Equilibrium, and Stability Dual Credit Physics Montwood High School R. Casao

2 Torque A force is needed to produce a change in rotational motion. The rate of change depends on the magnitude of the force and on the perpendicular distance of its line of action from the axis of rotation. The line of action of a force is an imaginary line extending through the force vector arrow (the line along which the force acts.

3 The figure shows that r  = r∙sin , where r is the straight line between the axis of rotation and the point at which the force and  is the angle between the line of r and the force F. The perpendicular distance r  is called the moment arm or lever arm or torque arm. Previously, I taught you to resolve the force into a component parallel to the axis of rotation and a component perpendicular to the axis of rotation. –The parallel force component passes through the pivot and does not produce a rotation. –The perpendicular force component produces the rotation of the object around the pivot point.

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5 Torque equation:  = r  ∙F = r∙F·sin  = r  ∙F  angle  is the angle between r and F Unit: m·N; this is the same as the unit of work and energy, N·m = J. The torque unit is generally written in reverse order m·N to avoid confusion. Torque is not work and the unit is not a Joule. Rotational acceleration is not always produced when a force acts on a stationary rigid body. –When the force acts through the axis of rotation so that  = 0º, the  = 0. –When the force acts perpendicular to r,  = 90º and the  is a maximum.

6 The angular acceleration depends on where a perpendicular force is applied and on the length of the lever arm. –The torque produced by the vector T 1 at radius R 1 is greater than the torque produced by vector T 2 at radius R 2. Torque in rotational motion is similar to force in translational motion. –An unbalanced or net force changes translational motion; an unbalanced or net torque changes rotational motion.

7 Torque is a vector; its direction is always perpendicular to the plane of the force and the moment arm vectors. –Torque is a vector cross product (the multiplcation of two vectors to produce a resultant third vector): –The direction is given by a right-hand rule. –The figure shows a particle at point A in the xy plane with a single force F acting on the particle

8 The particle’s position relative to the origin O is given by the position vector r. To find the direction of the torque, slide the force vector without changing its direction until its tail is at the origin O so that it is tail to tail with the position vector. Use the right hand rule to rotate the position vector r into the force vector F. Point fingers of right hand in direction of r and rotate fingers toward F. The thumb points in the direction of the torque.

9 Equilibrium Equilibrium refers to a condition in which the net forces acting on a system are balanced. –Unbalanced forces produce translational accelerations; balanced forces produce translational equilibrium. –Unbalanced torques produce rotational accelerations; balanced torques produce rotational equilibrium. Translational equilibrium occurs when the sum of the forces acting on a body is 0 N; the body remains at rest (static equilibrium) or in motion with a constant velocity (dynamic equilibrium). –Condition for translational equilibrium:  F = 0 N.

10 Forces with lines of action through the same point are called concurrent forces (a and b in the figure to the right). When the vector sum of concurrent forces is 0 N, the body is in translational equilibrium. In part c of the figure, the  F = 0 N, but the opposing forces will cause the object to rotate about the pivot; the object will not be in static equilibrium. A pair of equal and opposite forces that do not have the same line of action is called a couple. The condition  F = 0 N is a necessary but not sufficient condition for equilibrium.

11 Condition for rotational equilibrium:  = 0 m·N If the sum of the torques acting on an object is 0, the object is in rotational equilibrium and either remains at rest (no rotation) or rotates with a constant angular velocity  (  = 0 rad/s 2 ). A body is in mechanical equilibrium when the conditions for both translational and rotational equilibrium are satisfied:  F = 0 N and  = 0 m·N A rigid body in mechanical equilibrium may be either at rest or moving with a constant linear and/or angular velocity.

12 Static equilibrium is the condition that exists when a rigid body remains at rest; v = 0 m/s and  = 0 rad/s. Translational static equilibrium example 8.4 Torque is a vector and has a direction. –We can designate torque directions as positive and negative depending on the rotational acceleration they produce. –Rotational directions are taken as clockwise or counterclockwise around the axis of rotation. –A torque that produces a counterclockwise rotation is positive; a torque that produces clockwise rotation is negative.

13 Rotational static equilibrium Example 8.5 Static equilibrium Example 8.6 Stability and Center of Gravity The equilibrium of a particle or a rigid body can be stable or unstable in a gravitational field. For rigid bodies, the equilibrium is analyzed in terms of the center of gravity of the body. The center of gravity is the point at which all of the weight of an object can be considered to be acting. When the acceleration of gravity is constant, the center of gravity and the center of mass coincide.

14 If an object is in stable equilibrium, any small displacement results in a restoring force or torque which tends to return the object to its original equilibrium position. The ball in the bowl is in stable equilibrium. Any move- ment of the ball within the bowl returns the ball to the bottom of the bowl. The pointed object is also in stable equilibrium.

15 Any slight displacement of the boject raises its center of gravity and a restoring gravitational force tends to return the object to the position of minimal potential energy (with the base flat on the surface – the equilibrium position). The gravitational force actually produces a restoring torque that is due to a component of the weight that tends to rotate the object about a pivot point back to its original position. This will only work as long as the center of gravity remains to the left of the pivot point.

16 If an object is in unstable equilibrium, any small displacement results in a torque that tends to rotate the object farther away from its equilibrium position. The ball on top of the bowl is in unstable equilibrium. Any movement of the ball from the top of the bowl causes the ball to leave its equilibrium position, decreasing its potential energy. The pointed object is also in unstable equilibrium. Any displacement of the center of gravity from directly over the pivot will result in a decrease in the object’s potential energy.

17 An object is in stable equilibrium as long as its center of gravity after a small displacement still lies above and inside the object’s original base of support. There will always be a restoring gravitational torque. –The line of action of the weight intersects the original base of support.

18 When the center of gravity or center of mass falls outside the base of support, the object falls over because of a gravitational torque that rotates it away from the equilibrium position. –The line of action of the weight does not intersect the original base of support.

19 Rigid bodies with wide bases and low centers of gravity are most stable and least likely to tip over. Ex. High speed racing cars Rigid bodes with a narrow base and a higher center of gravity are less stable and more likely to tip over. Ex. SUV’s

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