Angular velocity and angular acceleration d θ CHAPTER 9 ds i ! r ROTATION i θ i • Angular velocity and angular acceleration ! equations of rotational motion The arc length moved by the ith element in a rotating • Torque and Moment of Inertia ! Newton’s 2nd Law for rotation rigid, non-deformable disk is: ds i = ri dθ • Determination of the Moment of Inertia where d θ is in radians. The angular velocity of the ! Parallel axis theorem ! Perpendicular axis theorem rotating disk is defined as: dθ ω = , • Rotational kinetic energy dt ! power and so the linear velocity of the i th element (in the • Rolling objects (with no slip) direction of the tangent) is: dsi vi = = riω, dt where r i is the distance 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 Earth 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 Linear motion 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 acceleration of 3.00 rad/s2. After 5 s, what are 2 2 = (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 accelerations 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 force produces change in wound onto a pulley or gear 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 mass 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 action 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] (scalar) 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 x=ℓ ⎡ 3⎤ 2 2 2 x I = ∫ x M dx = M ⎢ ⎥ ( ℓ) ( ℓ) 3 x=−ℓ ⎣⎢ ⎦⎥ −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 plane 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 π r 2 − r 2 ( 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 .