Physics 201: Lecture 19, Pg 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

Physics 201: Lecture 19, Pg 2 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? (B) its initial angular acceleration ? (C) initial linear acceleration of its free end ? L m

Physics 201: Lecture 19, Pg 3 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (B) its initial angular acceleration ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! L m mg  F = 0 occurs only at the hinge but  z = I  z = r F sin 90° at the center of mass and (I CM + m(L/2) 2 )  z = (L/2) mg and solve for  z

Physics 201: Lecture 19, Pg 4 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (C) initial linear acceleration of its free end ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! L m mg a =  L

Physics 201: Lecture 19, Pg 5 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and use the Work-Energy Theorem 2. The hinge changes everything! m mg W = mgh = ½ I  2 W = mgL/2 = ½ (I CM + m (L/2) 2 )  2 and solve for  mg L/2 L

Physics 201: Lecture 19, Pg 6 Connection with Center-of-mass motion l If an object of mass M is moving linearly at velocity V CM without rotating then its kinetic energy is l If the object is both moving linearly and rotating then If an object with moment of inertia I CM is rotating in place about its center of mass at angular velocity  then its kinetic energy is

Physics 201: Lecture 19, Pg 7 But what of Rolling Motion l Consider a cylinder rolling at a constant speed. l Contact point has zero velocity in the laboratory frame. l Center of wheel has a velocity of V CM l Top of wheel has a velocity of 2V CM V CM CM 2V CM

Physics 201: Lecture 19, Pg 8 Rolling Motion l Now consider a cylinder rolling at a constant speed. V CM CM The cylinder is rotating about CM and its CM is moving at constant speed (V CM ). Thus its total kinetic energy is given by :

Physics 201: Lecture 19, Pg 9 Motion l Again consider a cylinder rolling at a constant speed. V CM CM Sliding only CM Rotation only v t =  R CM 2V CM V CM Both with |v t | = |v CM |

Physics 201: Lecture 19, Pg 10 Concept question l For a hoop of mass M and radius R that is rolling without slipping, which is greater, its translational or its rotational kinetic energy? A. Translational energy is greater B. Rotational energy is greater C. They are equal D. The answer depends on the radius E. The answer depends on the mass

Physics 201: Lecture 19, Pg 11 Example : Rolling Motion l A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? l Use Work-Energy theorem M  h M v ? Disk has radius R M M M M M Mgh = ½ Mv 2 + ½ I CM  2 and v =  R Mgh = ½ Mv 2 + ½ (½ M R 2 )(v/R) 2 = ¾ Mv 2 v = 2(gh/3) ½

Physics 201: Lecture 19, Pg 12 Question: The Loop-the-Loop … with rolling l To complete the loop-the-loop, how does the height h compare with rolling as to that with frictionless sliding? A.h rolling > h sliding B.h rolling = h sliding C.h rolling < h sliding h ? R ball has mass m & r <<R U=mg2R

Physics 201: Lecture 19, Pg 13 Example: Loop-the-Loop with rolling l To complete the loop the loop, how high do we have to release a ball with radius r (r <<R) ? l Condition for completing the loop the loop: Circular motion at the top of the loop (a c = v 2 / R) l Use fact that E = U + K = constant (or work energy) h ? R 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 b =mgh U=mg2R

Physics 201: Lecture 19, Pg 14 Example: Loop-the-Loop with rolling Work energy (  K = W) l K f = K CM + K Rot = mg (h-2R) (at top) K f = ½ mv 2 + ½ 2/5 mr 2  2 = mg (h-2R) & v=  r l ½ mgR + 1/5 mgR = mg(h-2R)  h = 5/2R+1/5R h ? R ball has mass m & r <<R U b =mgh U=mg2R Just a little bit more….

Physics 201: Lecture 19, Pg 15 Example: Loop-the-Loop … Energy Conservation l If rolling then ball has both rotational and CM motion! l E= U + K CM + K Rot = constant = mgh (at top) E= mg2R + ½ mv 2 + ½ 2/5 mr 2  2 = mgh & v=  r l E= mg2R + ½ mgR + 1/5 m v 2 = mgh  h = 5/2R+1/5R h ? R ball has mass m & r <<R U b =mgh U=mg2R h is just a little bit more….

Physics 201: Lecture 19, Pg 16 Torque : Rolling Motion (center of mass point) l A solid cylinder, with mass m and radius R, is rolling without slipping down an inclined plane. What is its angular acceleration  ?  F x = m a cm = mg sin  - f s  m Ra cm = mgR sin  - Rf s  F y = 0 = N – mg cos   (not used)  = -R f s = I cm  3.  = - a cm / R   -mR 2  = mgR sin  + I    -mR 2  – I  = mgR sin   = -mgR sin  (mR 2 + I) = -2g sin  / 3R  mg N fsfs 

Physics 201: Lecture 19, Pg 17 Torque : Rolling Motion (contact point) l A solid cylinder, with mass m and radius R, is rolling without slipping down an inclined plane. What is its angular acceleration  ?  = -R mg sin  = I  2. I = mR 2 + I cm  = -mgR sin  (mR 2 + I cm )  = 2g sin  / 3R  mg N fsfs 

Physics 201: Lecture 19, Pg 18 Torque : Come back spool (contact point) l 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? mg N fsfs F  = -rF = I  2. I = mR 2 + I cm = 3/2 mR 2  = -rF  (mR 2 + I cm ) = -2Fr  / 3R mg N fsfs F  = 0 = I   = 0 mg N fsfs F  = rF = I  2. I = mR 2 + I cm  = 2Fr  / 3R

Physics 201: Lecture 19, Pg 19 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  s is the coefficient of static friction at the contact point, what is the maximum force that can be applied before slipping occurs? mg N fsfs F 1  F x = m a cm = F - f s 2.  F y = 0 = N – mg 3.  = -R f s = I cm   - a cm  - f s mR 2 / I cm 4. f s ≤  s N =  s mg (maximum) 5. a cm = -  R m a cm = f s mR 2 / I cm = F - f s f s (1+ mR 2 / I cm ) =  s mg (1+ mR 2 / I cm ) = F F =  s mg (1+ mR 2 / ( ½ mR 2 ) ) = 3  s mg

Physics 201: Lecture 19, Pg 20 Modified Atwood’s Machine (More torques) l Two blocks, as shown, are attached by a massless string which passes over a pulley with radius R and rotational inertia I = ½ MR 2. The string moves past the pulley without slipping. The surface of the table is frictionless.  What are the tensions in the strings ? m1gm1g N T T T1T1 T1T1 m2gm2g mass 1  F y = m 1 a y = -m 1 g +T 1 mass 2  F x = m 2 a x = –T pulley  = I  = R T 1 – R T a y = a x = -  R = a m 1 a = -m 1 g +T 1 m 2 a = –T -I a / R = R T 1 – R T

Physics 201: Lecture 19, Pg 21 Modified Atwood’s Machine (More torques) l 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? m1gm1g N T T T1T1 T1T1 m2gm2g 1. m 1 a = – m 1 g +T 1 2. m 2 a = –T 3. -I a / R = R T 1 – R T Solve for a -I a / R 2 = m 1 a +m 1 g + m 2 a a = – m 1 g / (m 1 + ½ M + m 2 ) T= - m 2 a T 1 = m 1 a + m 1 g

Physics 201: Lecture 19, Pg 22 Angular Momentum l We have shown that for a system of particles, momentum is conserved if l Are the rotational equivalents? angular momentum is conserved if

Physics 201: Lecture 19, Pg 23 For Thursday l Read all of chapter 10 (gyroscopes will not be tested) l HW9