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Section 10.6: Torque

Section 10.6: Torque. Translation-Rotation Analogues & Connections. Translation Rotation Displacement x θ Velocity v ω Acceleration a α Mass (moment of inertia) m I Kinetic Energy (K) (½)mv 2 (½)I ω 2

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Section 10.6: Torque

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  1. Section 10.6: Torque

  2. Translation-Rotation Analogues & Connections Translation Rotation Displacement x θ Velocity v ω Acceleration a α Mass (moment of inertia) m I Kinetic Energy (K) (½)mv2 (½)Iω2 Force F ? CONNECTIONS s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r

  3. Newton’s 1st Law(rotational language version): “A rotating body will continue to rotate at a constant angular velocity unless an external TORQUE acts.” • Clearly, to understand this, we need to define the concept of TORQUE. • Newton’s 2nd Law(rotational language version): Also needs torque.

  4. To cause an object to rotate requires a FORCE, F.(Cause of angular acceleration α). • BUT: The locationof the force on the object & the direction it acts are also important!  Introduce the torque concept. • Angular acceleration α F. • But alsoα (distance from the point of application ofFto the hinge = Moment Arm or Lever Arm,d)  From experiment! 

  5. Moment Arm d=distance of the axis of rotation from the “line of action” of force F • d=Distance which is to both the axis of rotation and to an imaginary line drawn along the direction of the force (“Line of Action”). • Line of Action Imaginary line extending out both ends of the force vector. Experiment finds that angular acceleration α (force) (moment arm) = Fd Define:TORQUE τ Fd τcausesα (Just as in the linear motion case,Fcausesa) Lower case Greek “tau”

  6. Newton’s laws & rotational motion We want to find a rotational analogue to force The figure is a top view of a door that is hinged on the left: The 4 pushing forces are of equal strength. Which of these will be the most effective at opening the door? • F1 will open the door. • F2will not. • F3 will open the door, but not as easily as F1. • F4 will open the door – it has same magnitude as F1, but we know it is not as effective as pushing at the outer edge of the door.

  7. Ability of force F to cause a rotation or twisting motion depends on 3 factors: 1.The magnitudeFof the force. 2.The distancerfrom point of application to pivot. 3.The angle at which F is applied. Make these idea quantitative. Figure is a force F applied at  one point on a rigid body. τ depends on the 3 properties, & is the rotational analogy to force.

  8. Let’s see this again

  9. Torque Units: Newton-meter = N m Sign convention: A torque that tends to rotate an object in a counterclockwise direction is positive. torque that tends to rotate an object in clockwise direction is positive. See Figure.

  10. Section 10.7: Rigid Object Under a Net Torque Consider the object shown. • The force F1 will tend to cause a counterclockwise rotation about O. • The force F2 will tend to cause a clockwise rotation about O • The net torque is: τnet = ∑τ = τ1+ τ2 = F1d1 – F2d2

  11. We’ve seen that toque is the rotational analogue to force. Now we need to learn what toque does. Figure is a model rocket engine attached to one end of lightweight, rigid rod. Tangential acceleration at& the angular acceleration α are related. Tangential component of force. Multiply both sides by r: Therefore, torque τcausesangular acceleration α. This relation is analogous to Newton’s 2nd Law F = ma

  12. Newton’s 2nd Law for Rotations Figure: A rigid body undergoes pure rotational motion about a fixed, frictionless, & unmoving axis. Net torque on the object is the sum of the torques on all the individual particles. Definition:Moment of Inertia Newton’s 2nd law for rotational motion

  13. Ex. 10.7: Net Torque on a Cylinder Double cylinder, shaped as shown. Attached to axle through center. Large part, radius R1 has rope around it & is pulled, with tension T1 to right. Small part, radius R2 has a rope around it & is pulled down with tension T2. (A) Net torque? ∑τ = τ1+ τ2 = T2R2 – T1R1 (B)T1 = 5 N, to R1 = 1 m, T2 = 15 N, R2 = 0.5 m. Net torque? Which direction will it rotate? ∑τ = (15)(0.5) – (1)(5) = 2.5 N m Direction is counterclockwise (positive torque)

  14. Ex. 10.8: Rotating Rod

  15. Ex. 10.10: Angular Acceleration of a Wheel Wheel, radius R, mass M, moment of inertia I. A cord is wrapped around it & attached to mass m. System is released & m falls & wheel rotates. Find the tension T in the cord, acceleration a of falling m, angular acceleration α of wheel. Newton’s 2nd Law for wheel: ∑τ = Iα(1) Tension T is tangential force on wheel & is the only force producing a torque. So:∑τ = TR (2) m moves in a straight line Newton’s 2nd Law for m: SFy = ma = mg – T (3) No slipping of cord. a = αR (4) From (1) & (2):α = (∑τ/I) = (TR)/I (5) From (4) & (5):a = (mg – T)/m = (TR2)/I (6) Solve for T from (6):T = (mg)/[1 + (mR2/I)] Put into (6) to get a:a = g/[1 + (I/mR2)] Put into (4) to get α:α = g/[R + (I/mR)]  a

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