Chapter 10: Rotational Motion Thursday March 5th • Review of rotational variables • Review of rotational kinematics equations • Rotational kinetic energy • Rotational inertia • Rolling motion as rotation and translation • Torque and Newton’s 2nd law (if time) • Examples, demonstrations and iclicker • Normal schedule after spring break, starting Monday 16th. • Material covered today relevant to LONCAPA due that day. • Normal lab schedule after spring break. • I will return mid-term exams on Tuesday after spring break. Reading: up to page 169 in Ch. 10

Review of rotational variables Angular position:

s θ= r

Angular displacement:

( in radians) Δθ = θ 2 − θ1

Average angular velocity: Units: rad.s-1, or s-1

ω avg

θ 2 − θ1 Δθ = = t2 − t1 Δt Δθ dθ ω = Lim = Δt→0 Δt dt

Instantaneous angular velocity: Average angular acceleration: Units: rad.s-2, or s-2

α avg

Instantaneous angular acceleration:

ω 2 − ω 1 Δω = = t2 − t1 Δt Δω dω α = Lim = Δt→0 Δt dt

Review of rotational kinematic equations

Equation number 10.7 10.8

Missing quantity

Equation

θ − θ0

ω = ω0 + αt

θ − θ 0 = ω 0t + α t 1 2

2

ω

10.9

ω = ω + 2α (θ − θ 0 )

t

10.6

θ − θ 0 = ω t = 12 (ω 0 + ω )t

α

θ − θ0 = ω t − α t

ω0

2

2 0

1 2

2

Important: equations apply ONLY if angular acceleration is constant.

Review: transforming rotational/linear variables Position:

s = θ r (θ in rads )

vt, at

ds dθ = r = ωr Tangential velocity: vt = dt dt

s r

θ

2π r 2π Time period for rotation: T = = v ω dvt dω = r = αr Tangential acceleration: at = dt dt v2 Centripetal acceleration: ar = = ω 2r r v depends on r

Kinetic energy of rotation

Consider a (rigid) system of rotating masses (same ω): ω 2 2 2 1 1 1 m3 K = m v + m v + m v + ⋅⋅⋅⋅⋅⋅⋅ 2 1 1 2 2 2 2 3 3 r3 m2 r 2 2 1 r1 = m v 2 i i m1 Rotation

∑

axis

where mi is the mass of the ith particle and vi is its speed. Re-writing this:

K = ∑ mi (ω ri ) = 1 2

2

1 2

(

)

2 2 m r ω ∑ ii

The quantity in parentheses tells us how mass is distributed about the axis of rotation. We call this quantity the rotational inertia (or moment of inertia) I of the body with respect to the axis of rotation.

I = ∑ mi ri2

K = 12 Iω 2

Calculating rotational inertia For a rigid system of discrete objects:

I = ∑ mi ri

∫

2

Therefore, for a continuous rigid object: I = r dm = 2

2 ρ r ∫ dV

• Finding the moments of inertia for various shapes becomes an exercise in volume integration. • You will not have to do such calculations.

r

axis

dm

• However, you will need to know how to calculate the moment of inertia of rigid systems of point masses. • You will be given the moments of inertia for various shapes.

Rotational Inertia for Various Objects

Parallel axis theorem • If you know the moment of inertia of an object about an axis though its center-of-mass (cm), then it is trivial to calculate the moment of inertia of this object about any parallel axis:

I PA

Icm com

dh Parallel rotation rotation axis axis

c.o.m. centeraxis of-mass axis

I PA = I cm + Md 2 • Here, Icm is the moment of inertia about an axis through the center-ofmass, and M is the total mass of the rigid object. • It is essential that these axes are parallel; as you can see from table 10-2, the moments of inertia can be different for different axes.

Rolling motion as rotation and translation

s =θR The wheel moves with speed ds/dt

⇒ vcm = ω R Another way to visualize the motion:

Rolling motion as rotation and translation

s =θR The wheel moves with speed ds/dt

⇒ vcm = ω R Another way to visualize the motion:

Rolling motion as rotation and translation

vcm = ω R Kinetic energy consists of rotational & translational terms:

K = I cmω + Mv 2

1 2

K= Modified mass:

1 2

{ fMR } R 2

2 cm

1 2

2 cm 2

v

= K r + Kt 2 cm

+ Mv

M' = (1+ f ) M

1 2

2 cm

= M'v 1 2

(look up f in Table 10.2)

Rolling Motion, Friction, & Conservation of Energy • Friction plays a crucial role in rolling motion (more on this later): • without friction a ball would simply slide without rotating; • Thus, friction is a necessary ingredient. • However, if an object rolls without slipping, mechanical energy is NOT lost as a result of frictional forces, which do NO work. • An object must slide/skid for the friction to do work. • Thus, if a ball rolls down a slope, the potential energy is converted to translational and rotational kinetic energy.

Ui

! α ! acm

Uf =Ui

ω v! K = Kt + Kr = Ui

cm

2 K = 12 I cmω 2 + 12 Mvcm