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CHAPTER-11 Rolling, Torque, and Angular Momentum

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Ch 11-2 Rolling as Translational and Rotation Combined Rolling Motion Rotation of a rigid body about an axis not fixed in space Smooth Rolling: Rolling motion without slipping Motion of com “O” and point “P” When the wheel rotates through angle , P moves through an arc length s given by s=R Differentiating with respect to t We get ds/dt= R d /dt v com = R

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Ch 11-2 Rolling as Translational and Rotation Combined Rolling motion of a rigid body: Purely rotational motion + Purely translational mption Pure rotational motion: all points move with same angular velocity . Points on the edge have velocity v com = R with v top = + v com and v bot = - v com Pure translational motion : All points on the wheel move towards right with same velocity v com

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Ch-11 Check Point 1 The rear wheel on a clowns’ bicycle has twice the radius of the front wheel. (a) When the bicycle is moving, is the linear speed at the very top of the rear wheel greater than, less than, or the same as that of the very top of the front wheel? (b) Is the angular speed of the rear wheel greater than, less than, or the same as that of the front wheel? 1. (a) v top-front =v top-rear =2 v com same; (b) v top-front = v top-rear = 2 front R front = 2 rear R rear rear / front = R front /R rear R rear = 2 R front rear / front = R front /R rear = 1/2 rear < front less

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Ch 11-3 Kinetic Energy of Rolling Rolling as a Pure Rotation about an axis through P Kinetic energy of rolling wheel rotating about an axis through P K= (I P 2 )/2 where I P = I com +MR 2 and R = v com K= (I P 2 )/2= (I com 2 +MR 2 2 )/2 K= (I com 2 )/2 + (Mv 2 com )/2 K= K Rot +K Trans

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Ch 11-4 The Forces of Rolling In smooth rolling, static frictional force f s opposes the sliding force at point P V com =R ; d/dt(V com )=d/dt(R ) a com =R d /dt=R Accelerating Torque acting clockwise; static frictional force f s tendency to rotate counter clockwise

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Ch 11-4-cont. Rolling Down a Ramp Rigid cylinder rolling down an incline plane, a com-x =? Components of force along the incline plane (upward) and perpendicular to plane Sliding force downward-static friction force upward; opposite trends f s -Mgsin =Ma com-x ; a com-x = (f s /M)-gsin To calculate f s apply Newtons Second Law for angular motion: Net torque= I Torque of f s about body com: f s R= I But =-a com-x /R; then f s =I com /R=-I com a com-x /R 2 a com-x =(f s /M)-gsin =(-I com a com-x /MR 2 )- gsin a com-x (1+I com /MR 2 ) = - gsin a com-x = - gsin /(1+I com /MR 2 )

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Ch-11 Check Point 2 Disk A and B are identical and rolls across a floor with equal speeds. The disk A rolls up an incline, reaching a maximum height h, and disk B moves up an incline that is identical except that is frictionless. Is the maximum height reached by disk B greater than, less than or equal to h? A is rolling and its kinetic energy before decent K A = I com 2 /2+ M(v com ) 2 /2 K B = M(v com ) 2 /2 v B

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Ch 11-5 The Yo-Yo Yo-Yo is Physics teaching Lab. Yo-Yo rolls down its string for a distance h and then climbs back up. During rolling down yo-yo loses potential energy (mgh) and gains translational kinetic energy (mv 2 com /2) and rotational kinetic energy ( I com 2 /2). As it climbs up it loses translational kinetic energy and gains potential energy. For yo-yo, equations of incline plane modify to =90 a com =- g/(1+I com /MR 0 2 )

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Ch 11-6 The Torque Revisited =r xF =r Fsin = r F = r F Vector product =r x F = i j k x y z F x F y F z

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The position vector r of a particle points along the positive direction of a z-axis. If the torque on the particle is (a) zero (b) in the negative direction of x and (c) in the negative direction of y, in what direction is the force producing the torque Ch-11 Check Point 3 =rxF=rfsin (a) =rfsin =0 ( =0, 180) (b)–i = k x F, i.e. F along j (c) –j=k x F i.e. F along -i

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Ch-11 Check Point 4 In part a of the figure, particles 1 and 2 move around point O in opposite directions, in circles with radii 2m and 4m. In part b, particles 3 and 4 travel in the same direction along straight lines at perpendicular distance of 4m and 2m from O. Particle 5 move directly away from O. All five particles have the same mass and same constant speed. (a) Rank the particles according to magnitude of their angular,momentum about point O, greatest first (b) which particles have negative angular momentum about point O. L = r mv r = 4m for 1 and 3 =2m for 2 and 4 =0 for 5 Ans: (a) 1 and 3 tie, then 2 and 4 tie, then 5 (zero); (b) 2 and 3

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Ch 11-7,8,9 Angular Momentum l =r x p =rp sin = r p = r p Newtons Second Law: F net = dp/dt; net = dl/dt For system of particles L= l i ; net = dL/dt

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Ch-11 Check Point 5 The figure shows the position vector r of a particle at a certain instant, and four choices for the directions of force that is to accelerate the particle. All four choice lie in the xy plane. (a) Rank the choices according to the magnitude of the time rate of change (dl/dt) they produce in the angular momentum f the particle about point O, greatest first (b) Which choice results in a negative rate of change about O? =(dl/dt)=rxF 1 = 3 = |rxF 1 |= |rxF 3 | and 2 = 4 =0

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Ch 11-7 Angular Momentum of a Rigid Body Rotating about a Fixed Axis Magnitude of angular momentum of mass m i l i = r i x p i =r i p i sin90= r i m i v i l i ( r i and p i ) Component of l i along Z-axis l iZ = l i sin =r i sin90 m i v i =r i m i v i v i = r i l iZ =r i m i v i =r i m i (r i )=r i 2 m i L z = l iZ = ( r i 2 m i ) =I (rigid body fixed axis)

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Ch-11 Check Point 6 In the figure, a disk, a hoop and a solid sphere are made to spin about fixed central axis (like a top) by means of strings wrapped around them, with the string producing the same constant tangential force F on all three objects. The three objects have the same mass and radius, and they are initially stationary. Rank the objects according to (a) angular momentum about their central axis (b) their angular speed, greatest first, when the string has been pulled for a certain time t. net =dl/dt=FR; l= net x t Since net =FR for all three objects, l hoop =l disk =l sphere f = i + t; net =I =FR; =FR/I i =0; f = i + t= t=FRt/I f = t=FRt/I I hoop =MR 2 ; I Disk =MR 2 /2; I sphere = 2/5 MR 2 f-hoop =FRt/I hoop =FRt/MR 2 f-Disk =FRt/I Disk =2(FRt/MR 2 ) f-Sphere =FRt/I Sphere =5(FRt/MR 2 )/2 Sphere, Disk and hoop angular speed

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Ch 11-11: Conservation of Angular momentum Newtons Second Law in angular form: net = dL/dt If net = 0 then L = a constant (isolated system) Law of conservation of angular momentum: L i = L I i i = I f f

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