Rolling Motion

Without friction, there would be no rolling motion. Assume: Rolling motion with no slipping  Can use static friction Rolling (of a wheel) involves: –The rotation about the Center of Mass (CM) PLUS –The translation of the CM

A wheel, moving on the ground with axle velocity v. The relation between the axle speed v & the angular speed ω of the wheel: v = r ω Rolls with no slipping!

Example Bicycle: v 0 = 8.4 m/s. Comes to rest after 115 m. Diameter = 0.68 m (r = 0.34m) a) ω 0 = (v 0 /r) = 24.7rad/s b) total θ = ( /r) = (115m)/(0.34m) = rad = 53.8 rev c) α = (ω 2 - ω 0 2 )/(2θ). Stopped  ω = 0  α = rad/s 2 d) t = (ω - ω 0 )/α. Stopped  ω = 0  t = 27.4 s r = 0.34m v 0 = 8.4 m/s  v = 0 d = 115m   v g = 8.4 m/s

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 Rotation Displacement x θ Velocity v ω Acceleration a α Force F τ (torque)

Section 10-4: 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 an object 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. But also α  (the distance from the point of application of F to the hinge  Lever Arm r  ) From experiment!

Lever Arm Angular acceleration α  force F, but also  distance from the point of application of F to the hinge (“Lever Arm”) Hinge F A = F B, but which gives a greater α ? R A, R B ≡ “Lever Arms” for F A & F B. α  “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  τ causes α (Just as in the linear motion case, F causes a) Lower case Greek “tau” 

Door Hinge The lever arm for F A is the distance from the knob to the hinge. The lever arm for F D is zero. The lever arm for F C is as shown. Forces at angles are less effective Torques: Due to F A : τ A = R A F A Due to F C : τ C = R C F C Due to F D : τ D = 0 (Since the lever arm is 0) τ C < τ A (For F C = F A ) R C is the Lever Arm for F C   R A  

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

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)

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

Example 10-7: Torque on a compound wheel 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 > R  = R B sin60º τ B = -R B F B sin60º τ A = R A F A τ = τ A + τ B = m N

Problem 25 τ 1 = - (0.24 m)(18 N) = m N τ 2 = +(0.24 m)(28 N) = 6.72 m N τ 3 = -(0.12 m)(35 N) = m N τ fr = m N Net torque: ∑τ = τ 1 + τ 2 + τ 3 + τ fr = -1.4 m N 35 N 28 N 18 N 24 cm 12 cm

Translational-Rotational Analogues & Connections Continue! Translation Rotation Displacementx θ Velocityvω Accelerationaα Force (Torque)Fτ Massm? CONNECTIONS v = rω a tan = rα a R = (v 2 /r) = ω 2 r τ = r  F