ch-10

Reference line (a) O P r (b) O P Reference line r _u s

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10.1

A compact disc rotating about a fixed axis through O perpendicular to the plane of the figure. (a) To define angular position for the disc, a fixed reference line is chosen. A particle at P is located at a radial distance r from the rotation axi s at O. (b) As the disc rotates, point P moves through an arc length s on a circular path of radius r. The angular position of P is .

Rotational Motion

Chapter 10

x y ,t_f ,ti r i O θ f θ

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10.2

A particle on a rotating rigid object moves from to along the arc of a circle. In the time interval t tf ti, the radial line of length r sweeps out an angle f i.

ω

ω

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10.3

The orange disk rotates in the directions indicated. The right-hand rule determines the direction of the angular velocity vector.

y P x O v r u s Figure 10.4

As a rigid object rotates about the fixed axis through O, the point has a tangential velocity that is always tangent to the circular path of radius r.

v :

x y O ar at P a

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10.5

As a rigid object rotates about a fixed axis through O, a particle at point P experiences a tangential component atand a radial component ar of translational acceleration. The total translational acceleration of this

particle is , where . a : r acrˆ a : :_a t :ar

vi mi r_i z axis O v

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10.6

A rigid object rotating about the z axis with angular speed . The kinetic energy of the particle of mass miis . The kinetic energy of the rigid object is called its rotational kinetic energy.

1 2mivi2

T

A B L E

10.2

Moments of Inertia of Homogeneous Rigid Objects With Different Geometries

Hoop or thin cylindrical shell I_CM= MR2 _R Solid cylinder or disk R ICM = 1₂MR2

Long thin rod with rotation axis through center ICM = 1₁₂ML2 L R Solid sphere ICM = 2₅MR2 Hollow cylinder R2 Long thin rod with rotation axis through end L Thin spherical shell I_CM= 2₃MR2 R1 ICM= 1₂M(R12 + R22) R Rectangular plate ICM= 1₁₂M(a2 + b2) b a I = 1₃ML2

L dr z r R

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10.8

(Example 10.4) The geometry for calculating the moment of inertia about the central axis of a uniform solid cylinder. m m M M O a a b b (b) y m m M M a a b b x (a)

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10.7

(Example 10.3) Four spheres form an unusual baton. (a) The baton is rotated about the y axis. (b) The baton is rotated about the z axis.

L

Pivot

M g

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10.9

(Example 10.5) A uniform rod rotates freely under the influence of gravity around a pivot at the left end.

r F sin F F cos d O Line of action φ φ φ φ r

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10.10

A force is applied to a wrench in an effort to loosen a bolt. The force has a greater rotating tendency about O as F increases and as the moment arm d increases. The component F sin tends to rotate the system about O. F : O d₂ d₁ F₂ F₁ Figure 10.11

The force tends to rotate the object counterclockwise about an axis through O, and tends to rotate the object clockwise.

F : 2 F : 1

O r P φ x F y τ = r F z r Figure 10.12

The torque vector lies in a direction perpendicular to the plane formed by the position vector and the applied force vector .:F

r : :

Right-hand rule – C = B × A C = A × B A θ B

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10.13

The vector product is a third vector having a magnitude AB sin equal to the area of the parallelogram shown. The vector is perpendicular to the plane formed by and , and its direction is determined by the right-hand rule.

B : A : C : C : A : B :

z x y R₁ R₂ O T₁ T₂

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10.14

(Example 10.6) A solid cylinder pivoted about the z axis through O. The moment arm of is R1, and the moment arm of T:2is R2.

T :

r F F O F (a) 2F (b) O 2r

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10.15

(a) The two forces acting on the object are equal in magnitude and opposite in direction. Because they also act along the same line of action, the net torque is zero and the object is in equilibrium. (b) Another situation in which two forces act on an object to produce zero net torque about O (but not zero net force).

53.0° 8.00 m (a) R T (c) 200 N 600 N (b) 53.0° 200 N 600 N 4.00 m 2.00 m R cosθ R sinθ T cos 53.0° T sin 53.0° θ θ θ

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10.16

(Example 10.8) (a) A uniform beam supported by a cable. A man walks out on the beam. (b) The free-body diagram for the beam–man system. (c) The free-body diagram with forces resolved into horizontal and vertical components.

(a) θ (b) θ mg O _s n P g f

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10.17

(Example 10.9) (a) A uniform ladder at rest, leaning against a frictionless wall. (b) The free-body diagram for the ladder.

T2 T1 (b) F M g m1 m1g T1 + m2g T2 + m₂ (a) m1 m2 + + M, R

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10.18

(Example 10.10) (a) An Atwood machine with a massive pulley. The pulley is modeled as a disk. (b) Free-body diagrams for the two hanging objects and the pulley.

φ O P r d d F θ s

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10.19

A rigid object rotates about an axis through O under the action of an external force applied at P.:F

M O R T m g m T

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10.20

(Example 10.11) An object hangs from a cord wrapped around a wheel. The tension in the cord produces a torque about the axle passing through O.

O z L = r × p m p φ y x r Figure 10.21

The angular momentum of a particle of mass m and linear momentum located at the position is given by . The value of depends on the origin about which it is measured and is a vector perpendicular to both and .:_r p:

L : L : r : :_p r : p : L :

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10.22

Angular momentum is conserved as Russian figure skater Evgeni Plushenko performs during the 2004 World Figure Skating Champion-ships. When his arms and legs are close to his body, his moment of inertia is small and his angular speed is large. To slow down for the finish of his spin, he moves his arms and legs outward, increasing his moment of inertia.

(© S tu ar t F ra nk lin /G et ty Im ag es )

(David Malin, Anglo-Australian Observatory)

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10.23

The Crab Nebula, in the constellation Taurus. This nebula is the remnant of a supernova explosion, which was seen on Earth in the year A.D. 1054. It is located some 6 300 lightyears away and is approximately 6 lightyears in diameter, still expanding outward.

v_i

m

F

O R

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10.24

(Example 10.13) When the string is pulled downward, the speed of the puck changes.

L_i Lf L CM O y z ∆L M g x n (a) (b) r

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10.25

Precessional motion of a top spinning about its symmetry axis. (a) The only external forces acting on the top are the normal force and the gravitational force . The direction of the angular momentum is along the axis of symmetry. The right-hand rule indicates that is in the xy plane. (b). The direction of is parallel to that of in part (a). That indicates that the top precesses about the z axis.

L : f :Li L: : L : : :_r :_F r : _{M g}: L : M g: n :

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10.26

Light sources at the center and rim of a rolling cylinder illustrate the different paths these points take. The center moves in a straight line (green line), whereas a point on the rim moves in the path of a cycloid (red curve). _{(Courtesy of Henry Leap and Jim Lehman)}

P CM

P′

2 v_CM

v_CM

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10.27

All points on a rolling object move in a direction perpendicular to a line through the instantaneous point of contact P. The center of the object moves with a velocity , whereas the point P moves with a velocity 2 v: .

CM v

:

h x v_CM ω M R θ Figure 10.28

(Example 10.15) A round object rolling down an incline. Mechanical energy of the object–surface–Earth system is conserved if no slipping occurs and there is no rolling resistance.

ChainSprocketCrank

m Figure P10.15 Figure P10.16 Problem 10.16 2h m1 I R m2 Figure P10.17

(John Lawrence/Stone/ Getty Images)

x O y = 3.00 m 4.00 kg 3.00 kg 2.00 kg y y = –2.00 m y = –4.00 m Figure P10.14

10.0 N 30.0° _a O b 12.0 N 9.00 N Figure P10.21 Fg1 Fg 2 2.00 m Figure P10.26 Figure P10.19 100 N 20.0° 20.0° 37.0° 2.00 m Figure P10.20

d P x O 2 m2 m1 CG Figure P10.27 Single point of contact 5.00 cm 30.0° 30.0 cm F Figure P10.29 d θ 2L Figure P10.31 10 000 kg (3 000 kg)g B A 2.00 m 6.00 m 1.00 m Figure P10.32

Figure P10.35 Figure P10.36 37.0° 15.0 kg T₁ m₁ 20.0 kg T₂ 2.00 m/s2 m2 Figure P10.37

Figure P10.44 Problems 10.44 and 10.50. I2 ωi ωf 1 r e t f A e r o f e B ω ω Figure P10.45 Figure P10.48 ) b ( ) a ( i ω ω_f

Figure P10.55 (b) (a) 1.50 m/s Figure P10.49 North Star (b) Cone of precession (a)

Figure P10.52 (a) At present, the spin axis of the

Earth points toward the North Star. (b) Torque on the spinning Earth will cause it to precess, so the spin axis will no longer be pointing in this direction in the future.