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d θ CHAPTER 9 ds i ! r i θ i • and ! equations of rotational The arc length moved by the ith element in a rotating • and of ! ’s 2nd Law for rotation rigid, non-deformable disk is: ds i = ri dθ • Determination of the where d θ is in . The angular velocity of the ! Parallel axis theorem ! Perpendicular axis theorem rotating disk is defined as: dθ ω = , • Rotational kinetic dt ! and so the linear velocity of the i th element (in the • objects (with no slip) direction of the tangent) is:

dsi vi = = riω, dt

where r i is the from the rotation axis. If the angular velocity changes there is angular acceleration ...

d θ ! ds ! a it DISCUSSION PROBLEM [9.1]: i a i ! ! r a i θ i ir You have a friend who lives in Minnesota, and you live ! ! a ir ⊥ a it in Florida . As the rotates, your linear velocity is ______hers, and your angular velocity is The angular acceleration of the disk is: ______hers. dω d dθ d2θ α = = . = , dt dt dt dt2 A: less than; equal to and the tangential acceleration of the i th element is: B: equal to; greater than dvi dω ait = = ri = riα. C: greater than; less than dt dt D: less than; greater than But, because the ith element is traveling in a circle, it E: greater than; equal to experiences a radial (centripetal) acceleration: 2 vi 2 air (= aic) = = riω . ri ! 2 2 The resultant linear acceleration is a i = air + ait . CONNECTION BETWEEN LINEAR Angular velocity ( ω) (vector) AND ROTATIONAL MOTION

1 Rotational motion Dimension: ω ⇒ [T] a ⇒ constant α ⇒ constant 1 [L] (Check: vi = riω ⇒ [L] = ). v = v! + at ω = ω! + αt [T] [T] 1 2 1 2 (x − x!) = v!t + at (θ − θ! ) = ω!t + αt Units: rad/s 2 2 2 2 2 2 v = v! + 2a(x − x!) ω = ω! + 2α(θ − θ!) Angular acceleration ( α) (vector)

You see, they’re 1 very similar Dimension: α ⇒ 2 [T]

Units: rad/s 2 r = 0.12 m: ω ! = 0: θ ! = 0: t = 5 s 3 00 rad/s2. r α = . ω = ?: θ = ?: a t = ?: a c = ? 2 (a) ω = ω! + αt = (3.00 rad/s )(5 s) = 15.0 rad/s.

Question 9.1: A disk of radius 12 cm, initially at rest, (b) v i = riω = (0.12 m)(15.0 rad/s) = 1.80 m/s (linear). begins rotating about its axis with a constant angular • tangential acceleration: a t = riα 2 2 2 acceleration of 3 .00 rad/s . After 5 s, what are = (0.12 m)(3.00 rad/s ) = 0.36 m/s . 2 • centripetal acceleration: a c = riω (a) the angular velocity of the disk, and = (0.12 m)(15.0 rad/s)2 = 27.0 m/s2. (b) the tangential and centripetal of a v2 (1.80 m/s)2 point on the perimeter of the disk? (Check ... a = = = 27.0 m/s2.) c r 0.12 m (c) How many revolutions were made by the disk in those 5 s? 1 2 (c) (θ − θ! ) = ω!t + αt 2 1 = (3.00 rad/s2 )(5 s)2 = 37.5 rad, 2 37.5 rad ⇒ n = = 5.97 rev. 2π In many applications a belt or chain is pulled from or As we saw in chapter 4 that produces change in wound onto a pulley or wheel ... motion. However, force does not always produce a change in rotational motion. It is torque that produces a v v t t change in rotational motion. Consider a m attached a t to a massless rigid rod that rotates around an axis O. The force F shown will cause the mass to rotate. R " F

θ As the string (chain or belt) is removed (or added), its r m O instantaneous velocity is the same as the tangential ℓ ℓ = r sin θ velocity at the rim of the wheel, providing there is no slip: The magnitude of the torque due to a force F on m is:

i.e., v t = Rω. τ = ℓF = (rsin θ)F, where ℓ is called the lever arm. The lever arm is the Also, under the same conditions, the instantaneous perpendicular distance from the axis of rotation (O) to acceleration of the string is the same as the tangential the line of of the force. acceleration at the rim of the wheel: " dvt dω NOTE: if F passes through O, i.e., ℓ = 0, then τ = 0 and i.e., at = = R = Rα. dt dt there will be no change in rotational motion. 2 F F sin " [M][L] [M][L] t = θ F Dimension: τ ⇒ [L] 2 = 2 (vector) [T] [T] θ F r = F cosθ Unit: N ⋅ m r m O ℓ Form the radial and tangential components of the force: Newton’s 2nd Law tells us that the tangential component of the force F t produces a tangential acceleration a t, " F F sin " F t = θ F i.e., F t = ma t. θ F r = F cosθ Therefore, the torque is r m r m τ = rF = mra , O O t t ℓ ℓ but, from earlier, the tangential (linear) acceleration is

related to the angular acceleration α, viz: a t = rα. ∴τ = mra = mr 2α. Then τ = ℓF = (rsin θ)F = r(Fsin θ) t i.e., τ = rFt. So torque (τ) produces angular acceleration (α). Note: the radial component F r, which passes through the axis of rotation, does not produce a torque and A rigid object that rotates about a fixed axis can be therefore it does not produce rotation; only the thought of as a collection of small, individual elements of mass m that each move in a circular path of radius r , tangential component F t produces a torque that results i i in rotation. where r i is measured from the axis of rotation. From above, for each separate mass element, we have Moment of inertia ... so 2 what’s that all about? τ i = miri α, where τ is the net torque r i i m i O on the ith element. Summing over all 2 Dimension: I ⇒ [M][L] () elements, the total net Units: kg ⋅ m2 torque on the object is: τ = ∑ τ = ∑ m r 2α = ∑ m r 2 α net i i i i i ( i i i ) Every object has a moment of inertia about an axis of = Iα, rotation. Its value depends not simply on mass but on 2 where I = ∑i miri is called the MOMENT OF INERTIA. how the mass is distributed around that axis. For a This is Newton’s 2nd Law for rotation, i.e., discrete collection of i objects, the moment of inertia τ net = Iα. about the rotation axis is: 2 I = ∑iIi = ∑i miri . “A net external torque acting on a body produces For a ‘continuous’ object: an angular acceleration, α, of that body given 2 2 I = Limit ∑i miri = ∫ r dm, by I α, where I is the moment of inertia.” mi→0 (viz: F net = ∑iFi = ma .) where m is a function of r. y

dx x x − ℓ ℓ 2 2 A rod is a ‘continuous’ object, so the moment of inertia is x=ℓ Question 9.2: Find the moment of inertia of a uniform 2 2 2 I = ∫ r dm = ∫ x dm. thin rod of length ℓ and mass M rotating about an axis x ℓ =− 2 perpendicular to the rod and through its center. The mass per unit length of the rod is M , so the mass of ℓ the small element of length dx is y dm = M dx. ( ℓ) x Substituting for dm, the integral becomes ℓ ℓ ℓ x= 2 ⎡ x3⎤ 2 I x2 M dx M ⎢ ⎥ = ∫ ( ℓ) = ( ℓ) x=−ℓ ⎣⎢ 3 ⎦⎥ ℓ 2 − 2 3 3 ⎡ ℓ (−ℓ )⎤ 1 2 = M − = Mℓ . ( ℓ)⎢ 24 24 ⎥ 12 ⎣ ⎦ [1] Show for yourselves that the moment of inertia of a rod of mass M and length ℓ about one end is y

dx x Question 9.3: Find the moment of inertia of the circular x 0 ℓ disk shown below, rotating about an axis perpendicular 1 2 to the and through its center. The mass of the disk I = Mℓ . 3 is 1.50 kg.

[2] Show for yourselves that the moment of inertia of a rod of mass M and length ℓ about an axis one-third the distance from one end is

y 10 cm 20 cm dx x x − ℓ 2ℓ 3 0 3

1 2 I = Mℓ . 9 r 2 2 1 The moment of inertia is given by I = ∫ r dm. Consider ∴I = M r 2 + r 2 2 ( 2 1 ) r1 1 a ring of radius r and = (1.50 kg) (0.20 m)2 + (0.10 m)2 2 ( ) dr width dr. If the mass of 3.75 10−2 kg m2. r the object is M, the = × ⋅ mass of the ring is 1 2πrdr [1] If r = 0, then I = MR2, dm = M , 1 disk 2 R r 2 r 2 π( 2 − 1 ) where R is the radius of the disk ( = r2). where r 1 and r 2 are the inner and outer radii of the object. Then, substituting for dm, r r 2 2 2πrdr 2M 2 3 I = ∫ r M = ∫ r dr [2] For a thin hoop, r 1 ≈ r2 = R, π r 2 − r 2 r 2 − r 2 R r1 ( 2 1 ) ( 2 1 ) r1 1 2 2 then Ihoop = M 2R = MR . 2 ( ) ⎡ 4 ⎤ r2 2M r M 4 4 = 2 2 ⎢ ⎥ = 2 2 (r2 − r1 ) r2 − r1 ⎣⎢ 4 ⎦⎥ 2 r2 − r1 ( ) r1 ( ) Providing the thicknesses of the disks are uniform, the M 1 moments of inertia do not depend on thickness. So, these = r 2 − r 2 r 2 + r 2 = M r 2 + r 2 . 2 r 2 r 2 ( 2 1 )( 2 1 ) 2 ( 2 1 ) expressions also apply to cylinders and tubes. ( 2 − 1 ) Sure, but, what’s the significance of I? Values of the moment of inertia for “simple” shapes ...

Thin rod Slab 1 2 1 I = ML I = M(a 2 + b2) 12 Remember, from chapter 4 ... 12 Mass ⇒ a measure of resistance to a change in b b linear motion, e.g., how difficult it is to start or L L a a stop linear motion. Slab 1 Thin rod I = Ma 2 1 3 I = ML2 Moment of Inertia ⇒ a measure of resistance to a 3 change in rotational motion, i.e., how difficult it is Hollow cylinder Hollow sphere 1 2 to start or stop rotational motion. I = M(R2 + r2) I = MR2 2 3 Thin-walled cylinder I = MR2

RMM04VD1.MOV

Solid cylinder Solid sphere Same 1 2 I = MR2 I = MR2 Different moments of inertia 2 5 O 5 kg b a O ′

The combined moment of inertia is I = Iruler + Imass, where I ruler and I mass are calculated about O − O ′ .

Since a >> b we assume the meter rule is a rod (viz: Question 9.4: A 1 m ruler has a mass of 0.25 kg. A 5 kg 1 2 mass is attached to the 100 cm end of the rule. What is Question 9.2) with I = Ma about O − O ′ . 3 its moment of inertia about the 0 cm end? 1 2 1 2 ∴Iruler = Ma = (0.25 kg)(1 m) 3 3 2 = 0.083 kg ⋅ m . The moment of inertia of the 5 kg mass about O − O ′ is 2 2 2 I mass = ma = (5.0 kg)(1 m) = 5.0 kg ⋅ m . Thus, the total moment of inertia of the ruler and mass is 2 I ruler + Imass = 5.083 kg ⋅ m . DISCUSSION PROBLEM [9.2]: Parallel axis theorem:

A pair of meter rulers are I cm Usually, the moment of inertia is I d placed so that their lower given for an axis that passes through ends are against a wall. the (cm) of the One of the rulers has a object. What if the object rotates large mass attached to its about an axis parallel to the axis upper end. If the meter through the center of mass for which rulers are released at the we don’t know the moment of inertia? The moment of same and allowed to inertia about a general (parallel) axis is given by: 2 fall, which one hits the I = Icm + Md floor first? where I cm is the moment of inertia about the center of mass, M is the mass of the object and d is the distance A: The meter ruler with the mass. between the parallel axes. B: The meter ruler without the mass. C: They hit the floor at the same time. Note: the two rotation axes must be parallel Perpendicular axis theorem:

z Consider a planar object (e.g., a thin Question 9.5: Four at the corners of a square disk or sheet) in the x,y plane. By with side length L = 2 m are conncted by massless rods. y y i definition, the moment of inertia x i r i The masses are m 1 = m3 = 3 kg and m 2 = m4 = 4 kg. mi about the z-axis (perpendicular to Find (a) the moment of inertia about the z-axis, (b) the x the plane of the object) is moment of inertia about an axis that is perpendicular to 2 2 2 2 2 I z = ∑i miri = ∑i mi(xi + yi ) = ∑i mixi + ∑i miyi the plane of the ensemble and passes through the center 2 But ∑ i mixi = Ix, i.e., the moment of inertia about x, and of mass of the system, (c) the moment of inertia about 2 the x-axis, which passes through m and m . ∑ i miyi = Iy, i.e., the moment of inertia about y. 3 4 y ∴Iz = Ix + Iy. L Note: the object must be planar m 1 m 2 Example of a disk: L

z y 1 2 m 4 m 3 Iz = MR . 2 x x z But, by , I x = Iy. 1 2 ∴Ix = Iy = MR . 4 y (a) Since we are dealing y 2 m 2 m 3 kg 4 kg with discrete masses 2 m 2 3 kg 4 kg I = ∑iIi = ∑i miri . cm 2 m Moment of inertia about 2 m D 4 kg 3 kg the z-axis: 4 kg 3 kg x I = m r 2 x z z ∑i i i 2 2 z = (3 kg)(2 m) + (4 kg)(2 2 m) 2 2 + (3 kg)(2 m) + (4 kg)(0) = 56 kg ⋅ m . (c) Since the ensemble is planar and confined to the x,y plane, we can use the perpendicular axis theorem, i.e., (b) By symmetry, the center of mass is at the center of I z = Ix + Iy. the square. But, by symmetry, I x = Iy. 2 2 2 ∴ Icm = ∑i miri = (3 kg)( 2 m) + (4 kg)( 2 m) 1 1 2 ∴Ix = Iz = 56 kg ⋅ m 2 2 2 2 2( ) + (3 kg)( 2 m) + (4 kg)( 2 m) = 28 kg ⋅ m . 2 • Check, using the parallel axis-theorem = 28 kg ⋅ m . 2 Check: I z = Icm + MD 2 2 2 2 2 2 I x = ∑i miri = (3 kg)(2 m) + (4 kg)(2 m) ∴Icm = Iz − MD = (56 kg ⋅ m ) − (14 kg)( 2 m) 2 2 = 28 kg ⋅ m . = 28 kg ⋅ m . (a) Using the parallel axis z theorem, we have for each ℓ Question 9.6: Four thin rods, each of length ℓ and mass rod ℓ M, are arranged to form a square, in the x,y plane, as y 2 I z = Icm + Md , shown. If the origin of the axes is at the center of the x where I is the moment of square, cm inertia through the center of mass of each rod and d = ℓ . (a) using the parallel axis theorem, show that 2 2 4 2 ⎛ 1 2 ℓ ⎞ 4 2 Iz = Mℓ . ∴Iz(total) = 4⎜ mℓ + m ⎟ = mℓ . 3 12 4 3 ⎝ ⎠ (b) Hence find I x and I y. (b) Since the object is planar we can use the perpendicular z ℓ axis theorem, i.e., ℓ Iz = Ix + Iy. y But because of symmetry I x = Iy. x 1 2 2 ∴Ix = Iy = Iz = mℓ . 2 3 A torque is required to rotate (or slow down) an object ... We saw in chapter 6 that a linearly moving object has but torque involves force ... translational ... an object rotating about an axis has rotational kinetic energy ... ! ! F F ! F r ds t γ If a rigid object is rotating d θ θ F r with angular velocity , v i ω O " γ + θ = 90 r th ω i m i the kinetic energy of the i F t = F cos γ O element is: 1 K = m v 2 i 2 i i When force is applied over a distance, is done, 1 2 2 = miri ω , since v i = riω. given by: 2 ! ! dW = F • ds = F.dscosγ = F cosγ.ds So, the total rotational kinetic energy is: = Ft.ds = Ft.rdθ = τ.dθ (J or N.m). 1 2 2 K = ∑iKi = ∑i miri ω Power is the at which the torque does work, 2 dW dθ 1 2 i.e., P = = τ = τω (). = Iω . dt dt 2 Note: P is the instantaneous power. 1 2 1 2 * Krot = Iω is the analog of Ktrans = mv . 2 2 • dW = τ.dθ is the analog of dW = F.ds. • P = τω is the analog of P = Fv. From earlier, power P = τω. Convert the torque to N ⋅ m and angular velocity to r ad/s, i.e., τ = 1.36 × 560 = 762 N ⋅ m, Question 9.7: The 3.9 liter V-8 fitted to a 488GTB and 2π(3700 rev/min) Ferrari develops 560 ft ⋅ lb of torque at 3000 rev/min. ω = = 387.5 rad/s. What is the power developed by the engine with these 60 s/min 5 parameters? (1 ft ⋅ lb = 1.36 N ⋅ m.) ∴ P = (762 N ⋅ m)(387.5 rad/s) = 2.95 ×10 watts. But 746 watts = 1 HP 2.95 ×105 watts ∴P = = 396 HP. 746 watts/HP (a) To find the rotational kinetic energy we need to know the angular velocity (ω). Treating the grindstone as a solid disk 1 1 I = MR2 = (16.0 kg)(0.50 m)2 = 2.0 kg ⋅ m2. 2 2 τ 10 N ⋅ m 2 Question 9.8: An electric motor exerts a constant torque Since τ = Iα, α = = 2 = 5.0 rad/s . I 2 kg ⋅ m of 10 .0 N ⋅ m to the shaft of a grindstone with mass The grindstone starts from rest ( ω ! = 0) so 16 .0 kg and radius 0.50 m. If the system starts from rest, t (5.0 rad/s2)(8.0 s) 40 rad/s. find ω = α = = 1 1 (a) the rotational kinetic energy of the grindstone ∴K = Iω2 = (2 kg ⋅ m2 )(40 rad/s)2 = 1600 J. 2 2 after 8.0 s,

(b) There are two ways to determine the work done by (b) the work done by the motor during this time, and the motor. (i) By the work-kinetic energy theorem we would (c) the average power delivered by the motor. expect the motor to have done 1600 J of work. (ii) We can use the expression W = τθ, but we need to find θ, i.e., the through which the grindstone has turned in 8.0 s. 1 2 1 2 2 θ = ω!t + αt = (5.0 rad/s )(8.0 s) = 160 rad 2 2 If a rigid object is suspended from an arbitrary point O and is free to rotate about that point, it will turn until the center of mass is vertically beneath the suspension point. ∴ W = τθ = (10.0 N ⋅ m)(160 rad) = 1600 J. y (c) The average power is the total work done divided by the total time interval, i.e., r cm cm ΔW 1600 J × P = = = 200 W. O x av x Δt 8.0 s cm Mg Note: the expression P = τω we derived earlier is actually the instantaneous power. We cannot use that expression here as ω is not constant. If the y-direction is vertical and the suspension point is not at the center of mass, the object will experience a net The instantaneous power actually increases linearly from torque given by zero at t = 0 to ( 10.0 N ⋅ m) × (40 rad/s) = 400 W at τ = Mgx cm, t = 8.0 s. where x cm is the x component of the center of mass. Therefore, the object will rotate until x cm = 0, i.e., the suspension point is direction above the center of mass. ℓ (a) Use Newton’s 2nd Law U1 = mg : K1 = 0 2 for rotation, i.e., I , x τ = α cm pivot where τ = mg ℓ and ℓ 2 mg 2 I 1 m 2. Then = ( 3) ℓ x 1 2 3g U2 = 0 : K2 = Iω α = τ = , 2 I 2ℓ which is the initial angular acceleration of the center of mass. The linear acceleration of the 100 cm end is then 3g 2 Question 9.9: The “zero” end of a 1 m ruler of mass a = ℓα = = 14.7 m/s , 2 0.25 kg is attached to a frictionless pivot, the other end is which is greater than g! Also note, the torque τ varies as free to rotate in the vertical plane. If the ruler is released the ruler swings down. from rest in the horizontal , what is

(b) To find the of the 100 cm end as it passes the (a) the initial acceleration of the 100 cm end of the vertical we use the conservation of , ruler, and i.e., U 1 + K1 = U2 + K2. (b) the linear speed of the 100 cm end as the ruler Taking the zero of the gravitational at the passes through the vertical? point where the center of mass is at its lowest point, then ℓ 1 mg + 0 = 0 + Iω2. 2 2 2v cm Consider a ball, cylinder, wheel ω v cm or disc) rolling on a surface without slipping. v = 0

• The point in contact with the surface has zero instantaneous velocity relative to the surface. mgℓ 3g ∴ω = = . • The velocity of the cm is v cm ( = rω = v), I ℓ • The velocity of a point at the is 2v cm (= 2v). The linear velocity of the 100 cm end as it passes the vertical is then Since the object has the same linear velocity as the cm, 2 v = ℓω = 3gℓ = 3(9.81 m/s )(1.0 m) i.e., v, the total kinetic energy is: = 5.42 m/s. 1 2 1 2 Ktot = mv + Iω 2 2 translational + rotational v where ω = r . So, as a ball rolls down a hill ...

v , ω Gravitational potential energy ⇒ translational energy + Consider rolling a sphere, cylinder and a hoop down an incline. Do they have the same velocity at the bottom?

v h

For each object, conservation of energy gives: RMA07VD2.MOV 1 1 mgh = mv2 + Iω2, 2 2gh 2 2 v = ⎛ I ⎞ 1+ ⎜ 2⎟ and with no slip v = Rω, where R is the radius of the ⎝ mR ⎠ 2 object. After manipulation we find: For a solid sphere I = mR 2. 5 2 2gh 2gh 10gh v = . ∴v2 = , i.e., v = . ⎛ I ⎞ 1+ 2 7 1+ ⎜ 2⎟ 5 ⎝ mR ⎠ ∴ For maximum velocity: I ⇒ smallest value. Since this is independent of m and R, a bowling ball and a pool ball would have the same at the bottom of So, in a race between objects rolling down a slope, the an incline, despite their different masses and radii! order would be (1) sphere, (2) cylinder, (3) hoop, and is completely independent of m and R! (a) Draw the free body diagram of all the acting on the ball. Use Newton’s 2nd Law down the incline:

Mg sin θ − f = Macm ... (i) N where a cm is the acceleration O of the center of mass. To get Question 9.10: A uniform solid ball of mass M and radius f R θ an expression for a cm, we take R rolls without slipping down an incline at an angle θ to torques about O, the center of the horizontal. Find (a) the frictional force acting at the Mg the ball. (Note: the normal point of contact with the surface, and (b) the acceleration force N and the force Mg do not contribute to the of the center of mass of the ball, in terms of M, R, g and θ. torque as their lines of action pass through O.) fR ∴ τ = fR = Iα, i.e., α = , I θ where I is the moment of inertia of the ball and α its angular acceleration. With no slip NOTE: there must be at the point of contact fR2 otherwise the ball would slide down the incline! acm = Rα = ...... (ii) I

Substituting for a cm in equation (i), we get fR2 MfR2 5 Mg sin θ − f = M = = f . ⎛ 2 ⎞ I ⎜ MR2⎟ 2 ⎝ 5 ⎠ 5 7 ∴Mg sin θ = f + f = f , 2 2 2 i.e., f = Mg sin θ. 7 Note that this is a static frictional force (because there is no slip).

(b) Substituting for f in equation (ii), we get fR2 ⎛ 2 ⎞ R2 5 acm = = ⎜ Mg sin θ⎟ = gsin θ. ⎝ ⎠ ⎛ 2 2⎞ I 7 ⎜ MR ⎟ 7 ⎝ 5 ⎠ Question 9.11: Suppose you can choose wheels of any design for a soapbox derby race car. If the total weight You can show yourselves that if we put I MR2, where = β of the is fixed, which type of wheel design should 1 for a hoop, 1 for a disk, etc., then a general β = β = 2 you choose if you want to have the best chance to win expression for the static frictional force acting on an the race? object rolling down an incline, with no slip, and the acceleration of the center of mass are: Mg sin θ gsin θ f = −1 and acm = . 1+ β 1+ β The total mechanical energy of the car is: translational kinetic energy + rotational kinetic energy. Conservation of energy gives: 1 2 1 2 Mgh = Ktrans + Krot = Mv + Iω , 2 2 F where M is the total mass of the car and I is the moment of h inertia of the wheels. With four wheels of radius r, for R example, we have, using the no-slip condition, 1 2 2 v 2 2 Krot = 4 × Iω = 2 βmr = 2βmv , Question 9.12: A cue ball is struck by a horizontal cue a 2 ( )( r) distance h above the center of the ball. If the cue ball is 1 2 2 ∴Mgh = Mv + 2βmv , to roll without slipping, what is h? Express your answer 2 2Mgh in terms of the radius R of the ball. i.e., v2 = . (M + 4βm) You can assume that the frictional force of the table on So, for maximum speed (v) at the bottom of the hill (and the ball is negligible compared with the applied force F. greater average speed) with M fixed, we want m and β to be as small as possible. Note: that the result does not depend on the radius of the wheels (r). So, solid wheels are a good choice with the mass concentrated as close to the axle as possible. F

h R f •

The net torque about the center is τ = Fh − fR. m 2 30 kg From earlier, if there is no-slip then v cm = Rω, 2 m 20 kg m 1 dv dω i.e., a = cm = R = Rα. cm dt dt Question 9.13: The 30 kg mass, shown above, is Using Newton’s 2nd Law F + f = macm. released from rest, from a distance of 2 m above the If F >> f , then F ≈ macm and τ ≈ Fh = Iα. ground. Modeling the pulley as a uniform disk with a F Fh I ∴ = acm = Rα = R , i.e., h = . radius of 10 cm and mass 5 kg, find (a) the speed of the m I mR 30 kg mass just before it strikes the ground, (b) the 2 For a solid sphere I = mR 2. angular velocity of the pulley at that instant, (c) the 5 tensions in the two strings, and (d) the time it takes for 2 ∴h = R. the 30 kg block to reach the ground. Assume the 5 2 bearings in the pulley are frictionless and there is no slip • h > R ⇒ top 5 between the string and the pulley. “SLIP” 2 • h < R ⇒ back spin 5 ω • • • • ω

m 30 kg m 2 1 m 2 30 kg m 1 v v v 20 kg m 2 m m 2 m v 1 2 20 kg m 1 m 2 We have both translational and rotational motion. The moment of inertia of the pulley is: i.e., 589 = 10v2 + 392 +15v2 +1.3v2 1 2 1 2 2 197 I = mR = (5 kg)(0.10 m) = 0.025 kg ⋅ m . ∴v = = 2.74 m/s. 2 2 26.3 (a) Using conservation of energy, as m 2 hits the ground 1 2 1 2 1 2 v 2.74 m2gh = m1v + m1gh + m2v + Iω . (b) ω = = = 27.4 rad/s. 2 2 2 R 0.10 v The angular velocity of the pulley ω = , so R T 1 T 2 (c) a Draw the free-body 2 1 2 a (30 kg)(9.81 m/s )(2 m) = (20 kg)v m 1g m 2g diagrams for the masses. 2 What is the upward acceleration of m 1? 2 1 2 +(20 kg)(9.81 m/s )(2 m) + (30 kg)v 2 2 2 v = v! + 2ah. 2 2 2 1 ⎛ v ⎞ v (2.74 m/s) 2 + (0.025 kg ⋅ m2)⎜ ⎟ , ∴a = = = 1.88 m/s . 2 ⎝ (0.10 m)⎠ 2h 2(2 m) Then T 1 − m1g = m1a, T 1 T 2 a i.e., T = m (g + a) = 234 N. a 1 1 m 1g m 2g Also T 2 − m2g = m2(−a), i.e., T 2 = m2(g − a) = 238 N.

1 2 (d) (y − y!) = h = at . 2 2h 2(2 m) ∴t = = 2 = 1.46 s. a 1.88 m/s