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Lecture 19

Lecture 19. Goals:. Specify rolling motion (center of mass velocity to angular velocity Compare kinetic and rotational energies with rolling Work combined force and torque problems Revisit vector cross product Introduce angular momentum. Connection with Center-of-mass motion.

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Lecture 19

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  1. Lecture 19 Goals: • Specify rolling motion (center of mass velocity to angular velocity • Compare kinetic and rotational energies with rolling • Work combined force and torque problems • Revisit vector cross product • Introduce angular momentum

  2. Connection with Center-of-mass motion • If an object of mass M is moving linearly at velocity VCMwithout rotating then its kinetic energy is • If an object with moment of inertia ICMis rotating in place about its center of mass at angular velocityw then its kinetic energy is • If the object is both moving linearly and rotating then

  3. But what of Rolling Motion 2VCM CM VCM • Consider a cylinder rolling at a constant speed. • Contact point has zero velocity in the laboratory frame. • Center of wheel has a velocity of VCM • Top of wheel has a velocity of 2VCM

  4. Rolling Motion • Now consider a cylinder rolling at a constant speed. VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

  5. Motion Both with |vt| = |vCM | Rotation only vt = wR Sliding only 2VCM VCM CM CM CM VCM • Again consider a cylinder rolling at a constant speed.

  6. Concept question • For a hoop of mass M and radius R that is rolling without slipping, which is greater, its translational or its rotational kinetic energy? • Translational energy is greater • Rotational energy is greater • They are equal • The answer depends on the radius • The answer depends on the mass

  7. Example :Rolling Motion Disk has radius R M M M M M h M v ? q M • A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? • Use Work-Energy theorem Mgh = ½ Mv2 + ½ ICMw2 and v =wR Mgh = ½ Mv2 + ½ (½ M R2)(v/R)2 = ¾ Mv2 v = 2(gh/3)½

  8. Question: The Loop-the-Loop … with rolling ball has mass m & r <<R U=mg2R h ? R • To complete the loop-the-loop, how does the height h compare with rolling as to that with frictionless sliding? • hrolling > hsliding • hrolling = hsliding • hrolling < hsliding

  9. Example: Loop-the-Loop with rolling • To complete the loop the loop, how high do we have to release a ball with radius r (r <<R) ? • Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) • Use fact that E = U + K = constant (or work energy) Ub=mgh ball has mass m & r <<R Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is U=mg2R h ? R

  10. Example: Loop-the-Loop with rolling • Work energy (DK = W) • Kf = KCM + KRot = mg (h-2R) (at top) • Kf = ½ mv2 + ½ 2/5 mr2w2 = mg (h-2R) & v=wr • ½ mgR + 1/5 mgR = mg(h-2R)  h = 5/2R+1/5R Ub=mgh ball has mass m & r <<R U=mg2R h ? R Just a little bit more….

  11. Torque :Rolling Motion (center of mass point) N mg q fs q • A solid cylinder, with mass m and radius R, is rolling without slipping down an inclined plane. • What is its angular acceleration a? 1. SFx = m acm = mg sin q - fs  m Racm = mgR sin q - Rfs SFy = 0 = N – mg cos q (not used) 2. St = -R fs = Icma 3. a = - acm / R  -mR2a= mgR sin q + Ia  -mR2a – Ia = mgR sin q a = -mgR sin q / (mR2 + I) = -2g sin q / 3R

  12. Torque :Rolling Motion (contact point) • A solid cylinder, with mass m and radius R, is rolling without slipping down an inclined plane. • What is its angular acceleration a? 1. St = -R mg sin q= I a 2. I = mR2 + Icm • = -mgR sin q / (mR2 + Icm) • = 2g sin q / 3R N mg q fs q

  13. Torque :Come back spool (contact point) N F mg fs N N F F mg mg fs fs • A solid cylinder, with mass m, inner radius r and outer radius R, being pulled by a horizontal force F at radius r? If the cylinder does not slip then what is the angular acceleration? 1. St = -rF= I a 2. I = mR2 + Icm = 3/2 mR2 • = -rF / (mR2 + Icm) = -2Fr/ 3R 1. St = rF= I a 2. I = mR2 + Icm • = 2Fr/ 3R 1. St = 0= I a • = 0

  14. Torque : Limit of Rolling (center of mass) • A solid cylinder, with mass m and radius R, being pulled by a horizontal force F on its axis of rotation. If ms is the coefficient of static friction at the contact point, what is the maximum force that can be applied before slipping occurs? 1. SFx = m acm = F - fs 2.SFy = 0 = N – mg 3.St = -R fs = Icma  - acm = - fs mR2/ Icm 4. fs≤ms N = ms mg (maximum) 5. acm = - a R m acm = fs mR2/ Icm = F - fs fs (1+ mR2/ Icm ) = ms mg (1+ mR2/ Icm ) = F F = ms mg (1+ mR2/ ( ½ mR2) ) = 3 ms mg N F mg fs

  15. Modified Atwood’s Machine (More torques) N T T T1 m2g T1 m1g • Two blocks, as shown, are attached by a massless string which passes over a pulley with radius R and rotational inertia I = ½ MR2. The string moves past the pulley without slipping. The surface of the table is frictionless. • What are the tensions in the strings ? mass 1 S Fy = m1 ay= -m1g +T1 mass 2 S Fx = m2 ax = –T pulley S t = Ia = R T1 – R T ay = ax= -aR = a m1 a= -m1g +T1 m2 a = –T -I a / R = R T1 – R T

  16. Modified Atwood’s Machine (More torques) N T T T1 m2g T1 m1g • Two blocks, as shown, are attached by a massless string which passes over a pulley with radius R and rotational inertia I. The string moves past the pulley without slipping. The surface of the table is frictionless. • What are the tensions in the strings? 1. m1 a= – m1g +T1 2. m2 a= –T 3. -I a / R = R T1 – R T Solve for a -I a / R2 = m1 a+m1g + m2 a a = – m1g / (m1 + ½ M + m2) T= - m2 a T1= m1 a+ m1g

  17. Angular Momentum • We have shown that for a system of particles, momentum is conserved if • Are the rotational equivalents? angular momentum is conserved if

  18. For Thursday • Read all of chapter 10 (gyroscopes will not be tested) • HW9

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