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 = rω

Rolls with no slipping!

ExampleBicycle: v0 = 8.4 m/s. Comes

to rest after 115 m. Diameter = 0.68 m (r = 0.34m)

a) ω0 = (v0/r) = 24.7rad/s

b) total θ = (/r) = (115m)/(0.34m)

= 338.2 rad = 53.8 rev

c) α = (ω2 - ω02)/(2θ). Stopped

ω = 0 α = 0.902 rad/s2

d) t = (ω - ω0)/α. Stopped

ω = 0 t = 27.4 s

r = 0.34mv0 = 8.4 m/s

v = 0d = 115m

vg = 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 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.

• 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 ArmAngular acceleration α force F, but also distance from the point of application of F to the hinge (“Lever Arm”)

Hinge

FA = FB, but which gives a greater α ?

RA, RB ≡ “Lever Arms” for FA & FB. 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 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.

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)

RC is the Lever Arm for FC

RA

In general, write

R= R sinθτ = RF sinθ

Units of τ: N m = m N

F= F sinθF = F cosθτ = RF sinθ

OR, resolve F into components F & F

τ = RF

These are the same of course!

Torque • In general, write

τ = RF

• 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 τnet = ∑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 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.

------------>R= RBsin60º

τB= -RBFBsin60ºτA= RAFA

τ = τA + τB = - 6.7 m N

Problem 25τ1 = - (0.24 m)(18 N)

= - 4.32 m N

τ2 = +(0.24 m)(28 N)

= 6.72 m N

τ3 = -(0.12 m)(35 N)

= - 4.2 m N

τfr = + 0.4 m N

Net torque:

∑τ = τ1 + τ2

+ τ3 + τfr

= -1.4 m N

35 N 28 N

18 N24 cm

12 cm

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) = ω2 r

τ = rF