 # Physics 101: Lecture 18, Pg 1 Physics 101: Lecture 18 Rotational Dynamics l Today’s lecture will cover Textbook Sections 9.4 - 9.6: è Quick review of last.

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Physics 101: Lecture 18, Pg 1 Physics 101: Lecture 18 Rotational Dynamics l Today’s lecture will cover Textbook Sections 9.4 - 9.6: è Quick review of last lecture: torque, center of gravity, equilibrium è Moment of inertia è Newton’s second law for rotational motion è Rotational work and kinetic energy è Angular momentum

Physics 101: Lecture 18, Pg 2 Torque and Equilibrium Mgmg Find total torque about this axis M paint g FAFA  (F A ) = 0  (Mg) = -Mgd 2  (paint) = -M paint gd 3  (mg) = mgd 1 Total torque = 0 = mgd 1 –Mgd 2 - M paint gd 3 Thus, when M paint =(md 1 -Md 2 )/d 3 then the plank is in equilibrium. d1d1 d2d2 d3d3 Find M paint so that plank is in equilibrium.

Physics 101: Lecture 18, Pg 3 Moment of Inertia l When torque is the analogue of force what is the analogue of mass or what is the measure of inertia of a rigid body in rotational motion about a fixed axis ? Consider a tangential force F T acting on a particle with mass M rotating on a circular path with radius R. The torque is given by  = F T R = M a T R = M  R R = (M R 2 )  I=M R 2 is called the moment of inertia of the particle. For any rigid body : I=  (m r 2 ) SI unit: [kg m 2 ] Any rigid body has an unique total mass, but the moment of inertia depends on how the mass is distributed with respect to the axis of rotation.

Physics 101: Lecture 18, Pg 4 Moments of Inertia of Common Objects Hollow cylinder or hoop about central axis I = MR 2 Solid cylinder or disk about central axis I = MR 2 /2 Solid sphere about center I = 2MR 2 /5 Uniform rod about center I = ML 2 /12 Uniform rod about end I = ML 2 /3

Physics 101: Lecture 18, Pg 5 Concept Question The picture below shows two different dumbbell shaped objects. Object A has two balls of mass m separated by a distance 2L, and object B has two balls of mass 2m separated by a distance L. Which of the objects has the largest moment of inertia for rotations around the x-axis? 1. A 2. B 3. Same CORRECT x 2L L m m 2m A B I = mL 2 + mL 2 = 2mL 2 I = 2m(L/2) 2 + 2m(L/2) 2 = mL 2

Physics 101: Lecture 18, Pg 6 Newton’s 2nd Law l If a net torque is applied to a rigid body rotating about a fixed axis, it will experience an angular acceleration:  ext = I    in rad/s 2 For the same net torque, the angular acceleration is the larger the smaller the moment of inertia.

Physics 101: Lecture 18, Pg 7 Rotational Work and Kinetic Energy Translational kinetic energy: KE trans = 1/2 m v 2 Rotational kinetic energy: KE rot = 1/2 I  2  in rad/s SI Unit: [J] Rotation plus translation: KE total = KE trans + KE rot = ½ m v 2 + 1/2 I  2 Work done by a constant torque in turning an object through an angle  : W R =    in rad SI Unit: [J]

Physics 101: Lecture 18, Pg 8 Conservation of Mechanical Energy l The total mechanical energy for a rigid body with mass m and moment of inertia I is given by the sum of translational and rotational kinetic energy and gravitational potential energy: E = KE trans + KE rot + E pot = = ½ m v 2 + ½ I  2 + m g h If the work done by non-conserving forces and torques is zero, the total mechanical energy is conserved (final equals initial total energy) : E f = E 0 if W nc =0

Physics 101: Lecture 18, Pg 9 Angular Momentum l The rotational analogue to momentum (1-dim: p = m v) in linear motion is angular momentum: L = I   in rad/s SI unit: [kg m 2 /s] Conservation of Angular Momentum: If the net average external torque is zero, the angular momentum is conserved, i.e. the final and initial angular momenta are the same : L f = L 0 if   ave,ext = 0

Physics 101: Lecture 18, Pg 10 Rotation Summary (with comparison to 1-d linear motion) Angular Linear See text: chapters 8-9 See Table 8.1, 9.2  = I  F = m a  = constant, t 0 =0sa = constant, t 0 =0 s  =  0 +  t v = v 0 + a t  =  0 +  0 t + ½  t 2 x = x 0 + v 0 t + ½ a t 2  2 =  0 2 + 2  (  -  0 ) v 2 = v 0 2 + 2 a (x-x 0 ) W =  W = F s L = I  p = m v

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