CT1:A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is

A. half the ladybug’s.B. the same as the ladybug’s.C. twice the ladybug’s.D. impossible to determine.

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.1 Angular position, displacement, velocity and acceleration

CCW +

Angular Position

Counterclockwise is positive

Radians

= s/r (a dimensionless ratio)

Angular Displacement

Average angular velocity

av = angular displacement / elapsed time

av = /t

Instantaneous angular velocity

= lim /t t 0

Angular VelocityCCW +

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.1 Angular position, displacement, velocity and acceleration

Average angular acceleration

av = angular velocity / elapsed time

av = /t

Instantaneous angular acceleration

= lim /t t 0

Angular AccelerationCCW +

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.1 Angular position, displacement, velocity and acceleration

CCW +

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.2 Rotational kinematics: The Rigid Object Under Constant Angular Acceleration

Equations for Constant Acceleration Only

1. vf = vi + axt f = i + t

2. xf = xi + (vi + vf) t / 2 f = i + (i + f) t / 2

3. xf = xi + vi t + axt2/2 f = i + i t + t2/2

4. vf2 = vi

2 + 2ax(xf – xi) f2 = i

2 + 2(f – i)

Assuming the initial conditions at t = 0

x = xi and = i

v = vi and = i

and a and are constant.

1. f = i + t

2. f = i + (i + f) t / 2

3. f = i + i t + t2/2

4. f2 = i

2 + 2(f – i)

P10.5 (p.300)

P10.6 (p.300)

CT2: Assume = 3 and remains constant. Which statement is always true.

A. at time 2t is twice at time t. B. at time 2t is three times at time t. C. at time 2t is one third at time t. D. the change in from 0 to 2t is twice

the change in from 0 to t. E. None of the above.

CT3: A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s linear speed is

A. half the ladybug’s.B. the same as the ladybug’s.C. twice the ladybug’s.D. impossible to determine.

CT4: A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down. At the instant shown in the figure, the radial component of the ladybug’s (Cartesian)acceleration is

A. in the +x direction.B. in the –x direction.C. in the +y direction.D. in the –y direction.E. in the +z direction.F. in the –z direction.G. zero.

CT5: A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down. At the instant shown in the figure, the tangential component of the ladybug’s (Cartesian) acceleration is

A. in the +x direction.B. in the –x direction.C. in the +y direction.D. in the –y direction.E. in the +z direction.F. in the –z direction.G. zero.

Connections

s = r

vt = r

at = r ar = v2/r P10.17 (p.301)

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.3 Angular and Translational Quantities

r

v

CT6: Assume 0. When = 0, then the total acceleration A. points inward. B. points outward. C. points along the tangent. D. there is not enough

information to tell.

About fixed axis O

I = miri2

K = I2/2 P10.17 (p.301)

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.4 Rotational Kinetic Energy

ri

vi

mi

O

P10.22 (p.301)

Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.5 Calculation of Moments of Inertia

Fig. 10.12, p.305

IO = ICM + MD2

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.6 Torque

CT7: You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut?

A B

C D

rsin

F┴ = 100cos33

33

d = 2sin57

57

= Fr = (100cos33N)(2m) = 168 Nm

= Fd = (100N)(2sin57m) = 168 Nm

= Frsin = (100N)(2m)(sin57) = 168 Nm

P10.32 (p.302)

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.7 The Rigid Object Under a Net Torque

= I P10.36 (p.303)

CT8: Assume remains constant. I f the moment of inertia is doubled, then

A. doubles.

B. halves.

C. also remains constant. D. there is not enough

information to tell.

CT9: Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations, how large must F2 be?

A. 0.25 NB. 0.5 NC. 1 ND. 2 NE. 4 N

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.8 Energy Considerations in Rotational Motion

W = = dK = mv2/2 + I2/2 for a system P10.37 (p.303)

P10.49 (p.304)

Rolling Without SlippingConstant v and

d = vt2r = vt

(2/t)r = v r = v

recall that r = vt

Ch 10 Rotation of a Rigid Object About a Fixed Axis10.9 Rolling Motion of a Rigid Object

K = Mvcm2/2 + Icm2/2 for rolling without slipping

scm = R vcm = R acm = R P10.56 (p.305)

CT10: A sphere (S), a cylinder C, and a hollow cylinder (HC) with the same outer radius and mass start from rest at the same position at the top of an incline. In what order do the three reach the bottom of the incline - which is first, second, and third? Think about I2/2 and energy conservation.

A. S-1st,C-2nd,HC-3rd

B. S-1st,HC-2nd,C-3rd

C. C-1st,HC-2nd,S-3rd

D. C-1st,S-2nd,HC-3rd

E. HC-1st,S-2nd,C-3rd

F. HC-1st,C-2nd,S-3rd

G. 3 way tie

CT11: Does the answer to the previous race depend on the radii of the objects?

A. Yes

B. No

CT12: Does the answer to the previous race depend on the masses of the objects?

A. Yes

B. No

Table 10.3, p.314