* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download V K M I + =
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Atomic theory wikipedia , lookup
Specific impulse wikipedia , lookup
Angular momentum operator wikipedia , lookup
Classical mechanics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Rotating locomotion in living systems wikipedia , lookup
Renormalization group wikipedia , lookup
Rolling resistance wikipedia , lookup
Equations of motion wikipedia , lookup
Work (physics) wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Mass versus weight wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Hunting oscillation wikipedia , lookup
Classical central-force problem wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Center of mass wikipedia , lookup
Seismometer wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Physics 201 – Lecture 19 Lecture 19 Example: Rotating Rod Goals: l • 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 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 1 Physics 201: Lecture 19, Pg 2 Example: Rotating Rod l Example: Rotating Rod 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 l (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! 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 L m Σ F = 0 occurs only at the hinge m but τz = I αz = r F sin 90° mg a=αL mg at the center of mass and (ICM + m(L/2)2) αz = (L/2) mg and solve for αz Physics 201: Lecture 19, Pg 3 Physics 201: Lecture 19, Pg 4 Connection with Center-of-mass motion 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 l (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! L m mg L/2 If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is 2 K T = 12 MVCM l W = mgh = ½ I ω2 If an object with moment of inertia ICM is rotating in place about its center of mass at angular velocity ω then its kinetic energy is K R = 12 I CMω 2 W = mgL/2 = ½ (ICM + m (L/2)2) ω2 and solve for ω l If the object is both moving linearly and rotating then 2 K = 12 I CMω 2 + 12 MVCM mg Physics 201: Lecture 19, Pg 5 Physics 201: Lecture 19, Pg 6 Page 1 Physics 201 – Lecture 19 But what of Rolling Motion l l l Now consider a cylinder rolling at a constant speed. VCM CM CM VCM The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by : 2 K TOT = 12 I CMω 2 + 12 MVCM Physics 201: Lecture 19, Pg 7 Physics 201: Lecture 19, Pg 8 Concept question Motion l Again consider a cylinder rolling at a constant speed. Both with |vt| = |vCM | Rotation only vt = ωR CM CM CM VCM 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 Rotational energy is greater They are equal The answer depends on the radius The answer depends on the mass Sliding only VCM 2VCM B. C. D. E. Physics 201: Lecture 19, Pg 9 Physics 201: Lecture 19, Pg 10 Question: The Loop-the-Loop … with rolling Example : Rolling Motion l l 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 l M M M To complete the loop-the-loop, how does the height h compare with rolling as to that with frictionless sliding? A. hrolling > hsliding B. hrolling = hsliding C.hrolling < hsliding Disk has radius R M h θ Mgh = ½ Mv2 + ½ ICM ω2 M l Rolling Motion 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 2VCM M l and M v? v =ωR ball has mass m & r <<R U=mg2R Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2 h? v = 2(gh/3)½ R Physics 201: Lecture 19, Pg 11 Physics 201: Lecture 19, Pg 12 Page 2 Physics 201 – Lecture 19 Example: Loop-the-Loop with rolling Example: Loop-the-Loop with rolling l l l Ub=mgh 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) ball has mass m & r <<R U=mg2R l h? R v Tangential 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 l l l Work energy (∆K = W) Kf = KCM + KRot = mg (h-2R) (at top) Kf = ½ mv2 + ½ 2/5 mr2 ω2 = mg (h-2R) & v=ωr ½ mgR + 1/5 mgR = mg(h-2R) h = 5/2R+1/5R Ub=mgh ball has mass m & r <<R U=mg2R v Tangential h? R = gR Just a little bit more…. Physics 201: Lecture 19, Pg 13 Physics 201: Lecture 19, Pg 14 Example: Loop-the-Loop … Energy Conservation l l l l If rolling then ball has both rotational and CM motion! E= U + KCM + KRot = constant = mgh (at top) E= mg2R + ½ mv2 + ½ 2/5 mr2 ω2 = mgh & v=ωr E= mg2R + ½ mgR + 1/5 m v2 = mgh h = 5/2R+1/5R Ub=mgh ball has mass m & r <<R U=mg2R v Tangential = gR h? R l l Torque : Rolling Motion (center of mass point) A solid cylinder, with mass m and radius R, is rolling without slipping down an inclined plane. What is its angular acceleration α? 1. ΣFx = m acm = mg sin θ - fs m Racm = mgR sin θ - Rfs mg θ -mR2 α= mgR sin θ + Iα -mR2 α – Iα = mgR sin θ α = -mgR sin θ / (mR2 + I) = -2g sin θ / 3R Physics 201: Lecture 19, Pg 16 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 α? 1. Στ = -R mg sin θ = I α 2. I = mR2 + Icm α = -mgR sin θ / (mR2 + Icm) α = 2g sin θ / 3R θ fs Physics 201: Lecture 19, Pg 15 l N ΣFy = 0 = N – mg cos θ (not used) 2. Στ = -R fs = Icm α 3. α = - acm / R h is just a little bit more…. l = gR l N θ Torque : Come back spool (contact point) 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. Στ = -rF = I α N 2. I = mR2 + Icm = 3/2 mR2 F α = -rF / (mR2 + Icm) = -2Fr / 3R mg fs mg fs N θ fs F mg 1. Στ = 0 = I α α= 0 Physics 201: Lecture 19, Pg 17 1. Στ = rF = I α 2. I = mR2 + Icm N α = 2Fr / 3R fs F mg Physics 201: Lecture 19, Pg 18 Page 3 Physics 201 – Lecture 19 l Modified Atwood’s Machine (More torques) 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 µs is the coefficient of static friction at the contact point, what is the maximum force that can be applied before slipping occurs? 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. v What are the tensions in the strings ? N l Σ Fy = m1 ay= -m1g +T1 T1 mass 2 T1 m1g Σ Fx = m2 ax = –T Σ τ = Iα = R T1 – R T m1 a = -m1g +T1 ay = ax= -αR = a m2 a = –T -I a / R = R T1 – R T Physics 201: Lecture 19, Pg 20 Modified Atwood’s Machine (More torques) Angular Momentum 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. v What are the tensions in the strings? N 2. m2 a = –T T1 3. -I a / R = R T1 – R T T1 m1g l -I a / m2g l = m1 a +m1g + m2 a is conserved if T1= m1 a + m1g r ∑ FExt r v = r dp =0 dt r τ = r × F = Iα r r r r L ≡ r × p = Iω r ∑ τ Ext r r v dL =r×F = =0 dt Physics 201: Lecture 19, Pg 22 For Thursday l Are the rotational equivalents? angular momentum Physics 201: Lecture 19, Pg 21 l r is conserved if a = – m1g / (m1 + ½ M + m2) T= - m2 a r r p ≡ mv momentum Solve for a R2 We have shown that for a system of particles, T T 1. m1 a = – m1g +T1 m2g pulley Physics 201: Lecture 19, Pg 19 l T T mass 1 N 1. ΣFx = m acm = F - fs F 2. ΣFy = 0 = N – mg mg 3. Στ = -R fs = Icm α fs - acm = - fs mR2 / Icm 4. fs ≤ µs N = µs mg (maximum) 5. acm = - α R m acm = fs mR2 / Icm = F - fs fs (1+ mR2 / Icm ) = µs mg (1+ mR2 / Icm ) = F F = µs mg (1+ mR2 / ( ½ mR2 ) ) = 3 µs mg Read all of chapter 10 (gyroscopes will not be tested) HW9 Physics 201: Lecture 19, Pg 23 Page 4