Section 10 6 Torque TranslationRotation Analogues Connections Translation

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)Iω2 Force F ? CONNECTIONS s = rθ, v = rω, at= rα, ac = (v 2/r) = ω2 r

• Newton’s 1 st 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 2 nd Law (rotational language version): Also needs torque.

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

• 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 Lower case Greek “tau” Define: TORQUE τ Fd τ causes α (Just as in the linear motion case, F causes a)

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? l l F 1 will open the door. F 2 will not. F 3 will open the door, but not as easily as F 1. F 4 will open the door – it has same magnitude as F 1, but we know it is not as effective as pushing at the outer edge of the door.

Ability of force F to cause a rotation or twisting motion depends on 3 factors: 1. The magnitude F of the force. 2. The distance r from 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.

Let’s see this again

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.

Section 10. 7: Rigid Object Under a Net Torque Consider the object shown. • The force F 1 will tend to cause a counterclockwise rotation about O. • The force F 2 will tend to cause a clockwise rotation about O • The net torque is: τnet = ∑τ = τ1+ τ2 = F 1 d 1 – F 2 d 2

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 τ causes angular acceleration α. This relation is analogous to Newton’s 2 nd Law F = ma

Newton’s 2 nd 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 2 nd law for rotational motion

Ex. 10. 7: Net Torque on a Cylinder Double cylinder, shaped as shown. Attached to axle through center. Large part, radius R 1 has rope around it & is pulled, with tension T 1 to right. Small part, radius R 2 has a rope around it & is pulled down with tension T 2. (A) Net torque? ∑τ = τ1+ τ2 = T 2 R 2 – T 1 R 1 (B) T 1 = 5 N, to R 1 = 1 m, T 2 = 15 N, R 2 = 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)

Ex. 10. 8: Rotating Rod

Ex. 10: Angular Acceleration of a 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 2 nd 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 2 nd 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 = (TR 2)/I (6) Solve for T from (6): T = (mg)/[1 + (m. R 2/I)] Put into (6) to get a: a = g/[1 + (I/m. R 2)] Put into (4) to get α: α = g/[R + (I/m. R)] a