If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced. Answer: make it twice as massive.

Test Your Understanding 10.3: Rigid-Body Rotation about a Moving Axis. A uniform cylinder rolls up an incline without slipping. As it rolls uphill, its speed decreases. The surface of the incline and the cylinder are both both perfectly rigid. Which of the forces on the cylinder exert(s) a torque about the center of the cylinder?

Answer: the friction force. Since the surface and incline are perfectly rigid, the normal force exerted by the incline on the cylinder is directed toward the cylinder's center. Hence, this force exerts zero torque about the center. The weight of the cylinder acts at its center (because the cylinder is uniform), so this force also exerts zero torque around the center. The friction force acts tangential to the circumference of the cylinder, so it does exert a torque around the center. Torque Magnitude Ranking Task. The wrench in the figure has six forces of equal magnitude. Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Rank from largest to smallest. smallest. To rank items as equivalent, overlap them.

We assume that all the forces are of the same magnitude. The moment arm values (ranked from largest to smallest): D, B and E, F, A, C. The moment arm that corresponds to F is greater than that for A because the 0 angle between the wrench axis and horizontal direction is less than 45 . Net Torque on a Pulley. The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored.

The statement that correctly describes the system shown in the figure: the angular acceleration of the pulley is nonzero. Indeed, block B moves with acceleration down, block A moves with acceleration up. Since the cord moves on the pulley without slipping, the linear speed of the pulley surface, v, increases, and this means that the d ω d ⎛ v ⎞ α = = ⎜ ⎟, angular acceleration is nonzero: dt dt ⎝ r ⎠ where r is the radius of the pulley. On the other hand, this also means that the net torque on the pulley is nonzero: τ α = where τ is the net torque on the pulley, I is the pulley’s moment of inertia. I Spinning Situations. Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above. For this problem, neglect any air resistance or friction between the merry-go-round and its foundation. Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merry-goround? It begins to rotate counterclockwise (as observed from above).

As long as there are no external torques acting on the system (which includes yourself, the merry-go-round, and the bicycle wheel) the angular momentum originally stored in the bicycle wheel is conserved. To stop the wheel from spinning (counterclockwise), you must exert a clockwise torque on it. By extending Newton's third law, this means that the wheel exerts a counterclockwise torque on you. If there is no friction between you and the merry-go-round, i.e., if the floor of the latter is completely smooth, this torque from the wheel will make you spin counterclockwise. However, if there is friction between your shoe soles and the floor, to prevent relative motion, this time, the floor exerts a clockwise torque on you, and you exert a counterclockwise torque on the floor of the merry-go-round. If the axle is completely smooth, then this torque will now make the merry-go-round spin counterclockwise. Of course, the ride will spin much slower than the wheel, because its moment of inertia is much larger.

Exercise 10.21. It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? IDENTIFY: SET UP: EXECUTE:

Apply Eq.

K=

1

2 Mv cm +

1

2 cm

I ω 2

2 2 For an object that is rolling without slipping, vcm = Rω . The fraction of the total kinetic energy that is rotational is

1 1 (1 2) I cmω 2 = = 2 2 2 2 (1 2 ) vcm + (1 2) I cmω 1(M/I cm )vcm /ω 1 + (MR 2 / I cm )

The moment of inertia of each object takes the form kinetic energy can be written as

(a) a uniform solid cylinder (b) a uniform sphere

cm

cm

1 1 + 1/ β

I= (1 2)

I= (2 5)

(c) a thin-walled hollow sphere

=

β 1 + β

2 . The ratio of rotational kinetic energy to total I= β MR

. The ratio increases as β increases.

2 , so the above ratio is 1 3. MR

2 so the above ratio is 2 7 . MR cm

I= (2 3)

2 MR so the ratio is 2 5 .

(d) a hollow, cylinder with outer radius R and inner radius R/ 2

cm

I= (5 8)

2 MR so the ratio is 5 13.

10.24. A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. How far up the smooth side will the marble go, measured vertically from the bottom? Express your answer in terms of h. How high would the marble go if both sides were as rough as the left side? IDENTIFY: SET UP:

Apply conservation of energy to the motion of the marble.

K=

1 2

mv2 +

1 2

ω I 2 , with

I=

2 5

MR2 . vcm = Rω for no slipping . Let y = 0 at the bottom of the bowl.

The marble at its initial and final locations is sketched in Figure 10.24. EXECUTE:

mgh =

(a) Motion from the release point to the bottom of the bowl: mgh =

1 2

mv 2 +

1 2

Iω 2.

2

1⎛ 2 10 ⎞⎛ v ⎞ mv2 + ⎜ mR 2 ⎟⎜ ⎟ and v = gh . 2 2⎝ 5 7 ⎠⎝ R ⎠

1

Motion along the smooth side: The rotational kinetic energy does not change, since there is no friction torque on 10 gh 5 1 v2 = 7 = h the marble, mv 2 + K rot = mgh′ + K rot . h′ = 2 2 g 2 g 7 (b) mgh = mgh′ so h′ = h . EVALUATE: (c) With friction on both halves, all the initial potential energy gets converted back to potential energy. Without friction on the right half some of the energy is still in rotational kinetic energy when the marble is at its maximum height.

Figure 10.24

Exercise 10.3 A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F 1 = 28.0 N, F2 = 14.3 N, and F3 = 17.0 N. The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise.

Answer: IDENTIFY and SET UP: Use Eq.(10.2) to calculate the magnitude of each torque and use the right-hand rule (Fig.10.4) to determine the direction. Consider Figure 10.3

Figure 10.3

Let counterclockwise be the positive sense of rotation. EXECUTE:

r1 = r2 = r 3 = (0.090 m) 2 + (0.090 m) 2 = 0.1273 m

τ 1 = − F1l 1

l1 = r 1 sin φ 1 = (0.1273 m) sin135° = 0.0900 m τ 1 = −(18.0 N)(0.0900 m) = −1.62 N ⋅ m τ 1 is directed into paper r

τ 2 = + F2 l 2

l2 = r 2 sin φ 2 = (0.1273 m) sin135° = 0.0900 m τ 2 = +(26.0 N)(0.0900 m) = +2.34 N ⋅ m τ 2 is directed out of paper r

τ 3 = + F3l 3

l3 = r 3 sin φ 3 = (0.1273 m)sin 90° = 0.1273 m τ 3 = +(14.0 N)(0.1273 m) = +1.78 N ⋅ m τ 3 is directed out of paper r

∑τ = τ

1

+ τ 2 + τ 3 = −1.62 N ⋅ m + 2.34 N ⋅ m +1.78 N ⋅ m = 2.50 N ⋅ m

EVALUATE: The net torque is positive, which means it tends to produce a counterclockwise rotation; the vector torque is directed out of the plane of the paper. In summing the torques it is important to include + or − signs to show direction.

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For the rod and axis at one end,

SET UP: EXECUTE:

EVALUATE:

α =

τ

=

I

Fl 1 3

2

Ml

=

I=

1 3

Ml2 .

3 F . Ml

Note that α decreases with the length of the rod, even though the torque increases.

Exercise 10.28 The engine of an aircraft propeller delivers an amount of power 179 hp to the propeller at a rotational velocity of 2500 rev/min. a) How much torque does the aircraft engine provide? b) How much work does the engine do in one revolution of the propeller? IDENTIFY: Apply P = τω and W = τΔθ . SET UP: P must be in watts, Δθ must be in radians, and ω must be in rad/s. 1 rev = 2π rad . 1 hp = 746 W .

π rad/s = 30 rev/min . EXECUTE:

(a) τ =

P

(175 hp )( 746 W / h p)

=

⎛ π rad/s ⎞ ( 2400 rev/min ) ⎜ ⎟ ⎝ 30 rev/min ⎠ (b) W = τΔθ = ( 519 N ⋅ m )( 2π rad ) = 3260 J ω

= 519 N ⋅ m.

EVALUATE: ω = 40 rev/s , so the time for one revolution is 0.025 s. P = 1.306 × 105 W , so in one revolution, W = Pt = 3260 J , which agrees with our previous result.

Problem 10.75: Rolling Stones A solid, uniform spherical boulder starts from rest and rolls down a 50.0 m high hill, as shown in the figure . The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill? IDENTIFY: boulder.

Apply conservation of energy to the motion of the

K=

SET UP:

1 2

mv2 +

without slipping. I= EXECUTE:

1 2

2 5

I 2 and v = Rω when there is rolling ω

mR2 .

Break into 2 parts, the rough and smooth sections.

Rough: mgh1 = 12 mv 2 + 12 Iω 2 . mgh1 = v2 =

2

1⎛ 2 ⎞⎛ v ⎞ mv2 + ⎜ mR 2 ⎟⎜ ⎟ . 2 2⎝ 5 ⎠⎝ R ⎠

1

10

gh1 . 7 Smooth: Rotational kinetic energy does not change.

mgh2 + vB =

10 7

gh1 + 2 gh2 =

10 7

1 2

mv2 + K rot =

1 2

2 + K rot . mvBottom

gh + 2

1 ⎛ 10

⎜ 2⎝ 7

⎞ ⎠

gh = 1⎟

1 2

2 B

v.

(9.80 m/s 2 )(25 m) + 2(9.80 m/s 2 )(25 m) = 29.0 m/s .

If all the hill was rough enough to cause rolling without slipping, vB =

10

g (50 m) = 26.5 m/s . A 7 smaller fraction of the initial gravitational potential energy goes into translational kinetic energy of the center of mass than if part of the hill is smooth. If the entire hill is smooth and the boulder slides without slipping, EVALUATE:

vB = 2 g (50 m) = 31.3 m/s . In this case all the initial gravitational potential energy goes into the kinetic energy of the translational motion.

Problem 10.87 A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. a) What is the final angular speed of the rod? Express your answer in terms of the variables v and L. b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision? IDENTIFY: Apply conservation of angular momentum to the collision. Linear momentum is not conserved because of the force applied to the rod at the axis. But since this external force acts at the axis, it produces no torque and angular momentum is conserved. SET UP: The system before and after the collision is sketched in Figure 10.87. EXECUTE: (a) mb = 14 mrod EXECUTE:

L1 =

1 8

L1 = mb vr = 14 mrod v( L / 2)

mrod vL

L2 = ( Irod + Ib )ω Irod =

1 3

mrod L2

Ib = mb r2 = 14 mrod ( L / 2) 2 Ib = 161 mrod L2 Figure 10.87

Thus L1 = L2 gives 1 8

v=

ω =

19 48

6 19

1 8

mrod vL = ( 13 mrod L2 + 161 mrod L2 )ω

Lω

v / L

(b) K1 =

1 2

mv2 =

1 8

mrod v2

K2 = 12 Iω 2 = 12 ( Irod + Ib )ω 2 = K2 = Then

1 2

( 1948) ( 196 )

K 2 K1

=

EVALUATE:

3 152 1 8

2

1 2

(

1 3

mrod L2 + 161 mrod L2 ) (6 v /19 L) 2

3 mrod v2 = 152 mrod v2

mrodv 2 mrod v2

= 3/19.

The collision is inelastic and K 2 < K 1.

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