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Physics 101: Lecture 18 Rotational Dynamics

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

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Physics 101: Lecture 18 Rotational Dynamics

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

  2. Find Mpaint so that plank is in equilibrium. Torque and Equilibrium Find total torqueabout this axis d1 d2 d3 FA Mpaintg mg Mg t(mg) = mgd1 Total torque = 0 = mgd1–Mgd2 - Mpaintgd3 t(FA) = 0 t(Mg) = -Mgd2 Thus, when Mpaint=(md1-Md2)/d3 then the plank is in equilibrium. t(paint) = -Mpaintgd3

  3. Moment of Inertia • 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 FT acting on a particle with mass M rotating on a circular path with radius R. The torque is given by t = FT R = M aT R = M a R R = (M R2) a I=M R2 is called the moment of inertia of the particle. For any rigid body : I= S (m r2) SI unit: [kg m2] 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.

  4. Moments of Inertia of Common Objects Hollow cylinder or hoop about central axis I = MR2 Solid cylinder or disk about central axis I = MR2/2 Solid sphere about center I = 2MR2/5 Uniform rod about center I = ML2/12 Uniform rod about end I = ML2/3

  5. CORRECT 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 m 2m 2L L x 2m m B A I = mL2 + mL2 = 2mL2 I = 2m(L/2)2 + 2m(L/2)2 = mL2

  6. Newton’s 2nd Law • If a net torque is applied to a rigid body rotating about a fixed axis, it will experience an angular acceleration: S text = I aa in rad/s2 For the same net torque, the angular acceleration is the larger the smaller the moment of inertia.

  7. Rotational Work and Kinetic Energy Work done by a constant torque in turning an object through an angle q : WR = tqq in rad SI Unit: [J] Translational kinetic energy: KEtrans = 1/2 m v2 Rotational kinetic energy: KErot = 1/2 I 2 w in rad/s SI Unit: [J] Rotation plus translation: KEtotal = KEtrans + KErot = ½ m v2+ 1/2 I 2

  8. Conservation of Mechanical Energy • 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 = KEtrans + KErot + E pot = = ½ m v2 + ½ I w2 + 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) : Ef = E0 if Wnc=0

  9. Angular Momentum • The rotational analogue to momentum (1-dim: p = m v) in linear motion is angular momentum: L = I ww in rad/s SI unit: [kg m2/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 : Lf = L0 if Stave,ext = 0

  10. Rotation Summary (with comparison to 1-d linear motion) See text: chapters 8-9 Angular Linear a = constant, t0=0s a = constant, t0 =0 s v = v0 + a t w = w0 + a t x = x0 + v0 t + ½ a t2 q = q0 + w0 t + ½ a t2 v2 = v02 + 2 a (x-x0) w2 = w02 + 2 a (q-q0) S t = I a S F = m a W = F s W = t q L = Iw p = m v See Table 8.1, 9.2

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