Rotational Dynamics Causes of rotational motion Analogies between

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.

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

Section 8 -4: Torque • 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 a body to rotate about an axis requires a FORCE, F. (Cause of angular acceleration α). • BUT: The location of the force on the body and the direction it acts are also important! Introduce the torque concept. • Angular acceleration α F. From experiment! • But also α (the distance from the point of application of F to the hinge = Lever Arm, r )

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

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 Lower case τ causes α Greek “tau” τ Fr (Just as in the linear motion case, F causes a)

Door Hinge Forces at angles are less effective Torques: Due to FA: r. A τ A = r AF A Due to FC : τC = r. CFC Due to FD: τD = 0 r. C is the Lever Arm for FC r. C (Since the lever arm is 0) τC < τA (For FC = FA) The lever arm for FA is the distance from the knob to the hinge. The lever arm for FD is zero. The lever arm for FC is as shown.

In general, write τ = r F These are the r = r sinθ τ = r. F sinθ OR, resolve F into components F & F same, of course! Units of τ: Nm=m. N τ = r. F F = F sinθ F = F cosθ τ = r. F sinθ

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

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

Exercise B

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

Example 8 -9 r = r. Bsin 60º -------> τB = -r. BFBsin 60º 2 thin disk-shaped wheels, radii r. A = 30 cm & r. B = 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 τ A= r AF A

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 35 N 12 cm 28 N 24 cm 18 N Net torque: ∑τ = τA + τB + τC + τfr = -1. 4 m N

Translational-Rotational Analogues & Connections Continue! Displacement Velocity Acceleration Force (Torque) Mass Translation x v a F m CONNECTIONS v = rω atan= rα a. R = (v 2/r) = ω2 r τ = r F Rotation θ ω α τ ?