Rotational Dynamics

2. Rotational Dynamics • Causes of rotational motion! • Analogies between linear & rotational motion continue. • Newton’s 3 Laws are still valid! But, here we write them using rotational language and notation.

3. Translational-Rotational Analogues Continue! ANALOGUES Translation Rotation Displacement x θ Velocity v ω Acceleration a α Force F τ (torque)

4. Section 8-4: Torque • 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.

5. To cause a body to rotate about an axis requires a FORCE, F.(Cause of angular acceleration α). • BUT: The locationof the force on the body and the direction it acts are also important!  Introduce the torque concept. • Angular acceleration α F. • But alsoα (the distance from the point of application of F to the hinge = Lever Arm, r)  From experiment! 

6. Angular acceleration α force F, but also  distance from the point of application of F to the hinge(“Lever Arm”) Lever Arm FA = FB, but which gives a greaterα ? Hinge RA, RB ≡ “Lever Arms” for FA & FB. α  “Lever Arm”

7. Section 8-4: Lever Arm • Lever Arm r = distance of the axis of rotation from the “line of action” of force F • r = Distance which is  to both the axis of rotation and to an imaginary line drawn along the direction of the force (“line of action”). • Find: Angular acceleration α (force) (lever arm) = Fr  Define:TORQUE τ Fr τcausesα (Just as in the linear motion case, Fcauses a) Lower case Greek “tau”

8. Forces at angles are less effective Torques: Due to FA: τA = rAFA Due to FC : τC = rCFC Due to FD: τD = 0 (Since the lever arm is 0) τC < τA (For FC = FA) Door Hinge   rA  rC is the Lever Arm for FC rC The lever arm for FA is the distance from the knob to the hinge. The lever arm for FDis zero. The lever arm for FC is as shown.

9. OR, resolve F into components F & F τ = rF     These are the same, of course!  In general, write τ = rF    F= F sinθ F= F cosθ τ = rF sinθ r= r sinθ τ = rF sinθ Units ofτ: N m = m N

10. Torque • In general, write τ = rF • Or, resolving F into components F|| and F: τ = rF • Even more generally: τ = rF sinθ • Units of torque: Newton-meters (N m)

11. Example 8-8:Biceps Torque τ = rF = 35 m N τ = rF = 30 m N

12. Exercise B

13. More than one torque? • If there is more than one torque: α τnet = ∑τ = sum of torques • Always use the following sign convention! Counterclockwise rotation  + torque Clockwise rotation  - torque

14. Example 8-9 r= rBsin60º --------------> τB = -rBFBsin60º 2 thin disk-shaped wheels, radii rA = 30 cm & rB = 50 cm, are attached to each other on an axle through the center of each. Calculate the net torque on this compound wheel due to the 2 forces shown, each of magnitude 50 N. τ= τA + τB = - 6.7 m N τA= rAFA

15. Problem 24 τA = - (0.24 m)(18 N) = - 4.32 m N τB = + (0.24 m)(28 N) = 6.72 m N τC = - (0.12 m)(35 N) = - 4.2 m N τfr = + 0.4 m N Net torque:∑τ = τA + τB + τC + τfr = -1.4 m N 35 N 28 N 12 cm 24 cm 18 N

16. Translational-Rotational Analogues & Connections Continue! Translation Rotation Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass m ? CONNECTIONS v = rω atan= rα aR = (v2/r) = ω2r τ = rF