FACULTY OF SCIENCE, AUTUMN 2007
PHYS 124 LEC A1 : Particles and Waves
(Instructor: Marc de Montigny)
Walker, Physics, Chapter 10: Rotational Kinematics and Energy
In this Chapter, we are mainly interested in the rotation of rigid
objects about fixed axes.
- Section 10-1: Angular Position, Velocity and Acceleration
- P. 282, Eq. 10-1: Angular Position θ (Definition)
- Angular displacement Δθ, angular velocity ω
and angular acceleration α are vectors.
- Bottom of P. 282: Sign Convention (θ > 0 counterclockwise,
θ < 0 clockwise)
- P. 283, Eq. 10-2: s = r θ (see Figure 10-2)
- P. 283, Eq. 10-3 and P. 284, Eq. 10-4: Angular Velocity (Definitions; analogous to linear
velocity) ω = Δθ/Δt (with Δt finite or infinitesimal)
- P. 284: Sign Convention for Angular Velocity
- P. 308, Problem 8
- P. 285, Eq. 10-5: Definition of Period T = 2π/ω. (When we will discuss
traveling waves, we will see again the very top of P. 285: In one revolution, we have
Δθ = 2π and Δt = T. As we will see, this implies
that Δθ/2π = Δt/T.)
- P. 285, Eqs. 10-6 and 10-7: Angular Acceleration (Definitions; analogous to linear
acceleration)
- Section 10-2: Rotational Kinematics
- P. 287, Table: Equations of Kinematics at Constant Angular
Acceleration
- Eq. 10-8: ω = ω0 + α t
- Eq. 10-9: θ = θ0 + ½(ω0 +
ω) t
- Eq. 10-10: θ = θ0 + ω0 t +
½α t²
- Eq. 10-11: ω² = ω0² + 2α
(θ - θ0)
- Not so surprisingly, they are similar to the equations encountered in
Chapter 2.
- P. 309, Problem 16
- P. 289, Active Example 10-1
- Section 10-3: Connection Between Linear and Rotational Quantities
- Eq. 10-12, P. 290 (vt = r ω) and
Eq. 10-14, P. 293 (at = r α) follow from one- and
second-order derivatives of Eq. 10-2, P. 282 (where r
is a constant), respectively.
- Question: What is the linear velocity (in m/s) of a point lying on the circumference
of an LP (i.e. Long-Play, or Vinyl) of diameter 30 cm that rotates at 33 rpm
(revolutions per minute)?
- T = 2πr/vt, that is, circumference divided by speed. Comparing
vt = (2π/T) r with vt = r ω, we conclude
that ω = 2π/T.
- P. 291, Eq. 10-13: acp = r ω²
- See middle of P. 293: Tangential Acceleration (due to changing angular
speed) Versus Centripetal Acceleration (due to changing direction). Since
at and acp are
perpendicular to one another, then the magnitude of the total acceleration
is given by a = (at² + acp²)½.
- P. 310, Problems 34, 35
- Section 10-4: Rolling Motion
- Have you ever realized that the bottom part of a rolling wheel is always
at rest with respect to the ground? This is because there is an
infinity of points around the wheel. Now, knowing that, and the
fact that, in general, μk < μs, what
is the most effective way to stop a car on an icy road, by slamming the
brakes, or by gently pumping them? Why?
- P. 295, Figure 10-10(a) and
Figure 10-11: Velocities in Rolling Motion
- P. 311, Problem 44
- Section 10-5: Rotational Kinetic Energy and the Moment of Inertia
- P. 296, Eq. 10-17: Rotational Kinetic Energy (K = ½ I ω².
Note the similarity with the analogous expression for the linear K.)
- The proof (P. 296) rests on K = ∑ (½ mi vi²),
with vi = ri ω, etc.
- P. 296, Eq. 10-18: Moment of Inertia
(I = ∑ mi ri²)
- Just like Mass measures the linear inertia of a body, the Moment of
inertia measures the rotational inertia of a body. (Reminder: go to
Section 5-1 of Chapter 5 for the concept of
Inertia.)
- Unlike mass, which depends on the composition of the object only, the
moment of inertia also depends on the axis of rotation
(via ri 's)
- P. 298, Table 10-1: Moments of Inertia for Various Objects and Axes
- Section 10-6: Conservation of Energy
- P. 300, Eq. 10-19: Kinetic Energy of a Rotating Solid Body
(K = ½ m v² + ½ I ω²) where v is the
speed of the centre of mass, and I is the moment of inertia about the centre
of mass of the object.
- P. 300, Eq. 10-20: Alternative Form
K = ½ m v² (1 + I/mr²)
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