Rotation Rolling Motion Chapter 10 Review Lana Sheridan De Anza College June 2, 2015 Overview • rolling motion • Chapter 10 practice Energy In total then, W = ∆Ktrans + ∆Krot + ∆U + ∆Eint where • ∆Ktrans represents the kinetic energy of the CM motion, and • ∆Krot is the rotational kinetic energy Rolling Motion e translation Combination of translation and rotation A combination of translation and rotation. v CM CM v ! v CM " R v ! 2v CM CM v CM v ! v CM Figure 10 P v CM P v !0 c The CM is translated at velocity vCM . rolling obj a combina and pure r Notice that instantaneously P is the pivot point of the rotation. Henry Leap and Ji Rolling Motion The pivot point of the rotation changes as different parts of the wheel contact the surface. ranslational speed of (10.28) holds whenever a cylpure rolling motion. mass for pure rolling (10.29) at speed v CM, staying mass of the object. As R u s s !R u Figure 10.24 For pure rolling motion, as the cylinder rotates ds through an angle center vCM = u=itsRω moves a linear distance s 5 R u. dt Kinetic Energy of a Rolling Object The kinetic energy of a rolling object is just the sum of the rotational KE and translational KE. Can see this by considering just an instantaneous rotation about the point P: K = 1 IP ω2 2 Using the parallel axis theorem: IP = ICM + mR 2 . K 1 (ICM + mR 2 )ω2 2 1 1 2 = ICM ω2 + mvCM 2 2 = KCM,rot + KCM,trans = Equation 10.30 to obtain Rolling down an incline K5 A sphere starts from rest at the top of an incline andv CM rolls down. Using 5 Rv, this equation ca Find the (translational) velocity of the center of mass at the K Total kinetic energy of a rolling object bottom of the incline. 1 2 M R h x u v S vCM Figure 10.26 A sphere rolling down an incline. Mechanical energy of the sphere–Earth system is conserved if no slipping occurs. The term 2 ICMv represents the center of mass, and the term 12Mv have if it were just translating th kinetic energy of a rolling objec the center of mass and the trans statement is consistent with the that the velocity of a point on th mass and the tangential velocity Energy methods can be used ing motion of an object on a ro which shows a sphere rolling wit top of the incline. Accelerated is present between the sphere a center of mass. Despite the pre occurs because the contact poin (On the other hand, if the sphe incline–Earth system would dec friction.) In reality, rolling friction cau energy. Rolling friction is due to For example, automobile tires fl Equation 10.30 to obtain Rolling down an incline K5 A sphere starts from rest at the top of an incline andv CM rolls down. Using 5 Rv, this equation ca Find the (translational) velocity of the center of mass at the K Total kinetic energy of a rolling object bottom of the incline. 1 2 The term 2 ICMv represents the center of mass, and the term 12Mv have if it were just translating th R kinetic energy of a rolling objec the center of mass and the trans h statement is consistent with the v x that the velocity of a point on th S vCM mass and the tangential velocity u Energy methods can be used ing motion of an object on a ro which shows a sphere rolling wit Figure 10.26 A sphere rolltop of the incline. Accelerated ing down an incline. Mechanical ∆K + ∆U = 0 is present between the sphere a energy of the sphere–Earth system is conserved if no slipping occurs. center of mass. Despite the pre occurs because the contact poin (On the other hand, if the sphe incline–Earth system would dec friction.) In reality, rolling friction cau energy. Rolling friction is due to For example, automobile tires fl M Equation 10.30 to obtain Rolling down an incline K5 A sphere starts from rest at the top of an incline andv CM rolls down. Using 5 Rv, this equation ca Find the (translational) velocity of the center of mass at the K Total kinetic energy of a rolling object bottom of the incline. 1 2 The term 2 ICMv represents the center of mass, and the term 12Mv have if it were just translating th R kinetic energy of a rolling objec the center of mass and the trans h statement is consistent with the v x that the velocity of a point on th S vCM mass and the tangential velocity u Energy methods can be used ing motion of an object on a ro which shows a sphere rolling wit Figure 10.26 A sphere rolltop of the incline. Accelerated ing down an incline. Mechanical ∆K + ∆U = 0 system is present between the sphere a energy of the sphere–Earth 1 1 center of mass. Despite the pre 2 is conserved 2 if no slipping occurs. ICM ω + MvCM − 0 + (0 − Mgh) = 0 occurs because the contact poin 2 2 (On the other hand, if the sphe incline–Earth system would dec friction.) In reality, rolling friction cau energy. Rolling friction is due to For example, automobile tires fl M Equation 10.30 to obtain Rolling down an incline K5 A sphere starts from rest at the top of an incline andv CM rolls down. Using 5 Rv, this equation ca Find the (translational) velocity of the center of mass at the K Total kinetic energy of a rolling object bottom of the incline. 1 2 The term 2 ICMv represents the center of mass, and the term 12Mv have if it were just translating th R kinetic energy of a rolling objec the center of mass and the trans h statement is consistent with the v x that the velocity of a point on th S vCM mass and the tangential velocity u Energy methods can be used ing motion of an object on a ro which shows a sphere rolling wit Figure 10.26 A sphere rolltop of the incline. Accelerated ing down an incline. Mechanical ∆K + ∆U = 0 system is present between the sphere a energy of the sphere–Earth 1 1 center of mass. Despite the pre 2 is conserved 2 if no slipping occurs. ICM ω + MvCM − 0 + (0 − Mgh) = 0 occurs because the contact poin 2 2 (On the other hand, if the sphe incline–Earth system would dec s friction.) 2gh In reality, rolling friction cau vCM = 1 + (ICM /MR 2 )) energy. Rolling friction is due to For example, automobile tires fl M Question Quick Quiz 10.71 A ball rolls without slipping down incline A, starting from rest. At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless. Which arrives at the bottom first? (A) The ball arrives first. (B) The box arrives first. (C) Both arrive at the same time. (D) It is impossible to determine. 1 Serway & Jewett, page 318. Question Quick Quiz 10.71 A ball rolls without slipping down incline A, starting from rest. At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless. Which arrives at the bottom first? (A) The ball arrives first. (B) The box arrives first. ← (C) Both arrive at the same time. (D) It is impossible to determine. 1 Serway & Jewett, page 318. ch as a marble and a croquet ball. phere, a solid cylinder, or a hoop, we would obtain Example 10.14 - Pulling a Spool fer. The constant factors that appear in the exprescylindrically symmetric of mass m and ut the Acenter of mass for the spool specific object. In radius all R sits at rest on acceleration a horizontal table withhave friction. pull on on a light string alue the would if theYou incline were wrapped around the axle (radius r ) of the spool with a constant horizontal force of magnitude T to the right. As a result, the spool rolls without slipping a distance L along the table with no rolling friction. Find the final translational speed of the center of mass of the spool. on a horizontal rapped around al force of magg a distance L L R S r spool. Figure 10.27 (Example 10.14) T ss than g sin u, the value the acceleration would have if the incline were Example 10.14 - Pulling a Spool AM adius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of magolls without slipping a distance L L R S T r nter of mass of the spool. Can use: W = ∆K . he motion of the spool when you distance L, notice that your hand ent from L. Figure 10.27 (Example 10.14) A spool rests on a horizontal table. A string is wrapped around the axle and is pulled to the right by a hand. continued n work–energy relationships for introductory physics,” The Physics Teacher, 43:10, 2005. ss than g sin u, the value the acceleration would have if the incline were Example 10.14 - Pulling a Spool AM adius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of magolls without slipping a distance L L R S r T nter of mass of the spool. Can use: W = ∆K . he motion of the spool when you distance L, notice that your hand ent from L. Figure 10.27 (Example 10.14) A spool rests on a horizontal table. A string is wrapped around the axle and is pulled to the right by a hand. r W =T L+L continued R n work–energy relationships for introductory physics,” The Physics Teacher, 43:10, 2005. ss than g sin u, the value the acceleration would have if the incline were Example 10.14 - Pulling a Spool AM adius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of magolls without slipping a distance L L R S r T nter of mass of the spool. Can use: W = ∆K . he motion of the spool when you distance L, notice that your hand ent from L. Figure 10.27 (Example 10.14) A spool rests on a horizontal table. A string is wrapped around the axle and is pulled to the right by a hand. r W =T L+L continued R s n work–energy relationships for introductory physics,” The Physics Teacher, 43:10, 2005. 2TL (1 + r /R) vCM = m + I/R 2 mine the moment of inertia of about axis. and Examplethis- structure Moment of this Inertia Section 10.7 Rotational Page 328, #44 Kinetic Energy Axis of Rotational rotation x KE Figure P10.43 44. Rigid rods of negligible mass lying along the y axis conW nect three particles (Fig. P10.44). The system rotates Q/C about the x axis with an y angular speed of 2.00 rad/s. Find (a) the moment of iner- 4.00 kg y ! 3.00 m tia about the x axis, (b) the total rotational kinetic energy x evaluated from 12Iv 2, (c) the O tangential speed of each 2.00 kg y ! "2.00 m particle, and (d) the total kinetic energy evaluated from 3.00 kg y ! "4.00 m 1 2 m v . (e) Compare the a2 i i answers for kinetic energy in Figure P10.44 parts (a) and (b). 45. The four particles in Figure P10.45 are connected by W rigid rods of negligible mass. The origin is at the cen- u s a p e p p in s s c t w c s Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? Ix = 92 kg m2 (b) & (d) Kinetic energy? Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? Ix = 92 kg m2 (b) & (d) Kinetic energy? K = 184 J (c) tangential speeds? Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? Ix = 92 kg m2 (b) & (d) Kinetic energy? K = 184 J (c) tangential speeds? v = rω Assuming pg333, the board# is 81 1.00 Example:(b)Rolling, m long and is supported at this limiting angle, show that the cup must be 18.4 cm from the moving end. 81. A uniform solid sphere of radius r is placed on the S inside surface of a hemispherical bowl with radius R. The sphere is released from rest at an angle u to the vertical and rolls without slipping (Fig. P10.81). Determine the angular speed of the sphere when it reaches the bottom of the bowl. r u R Figure P10.81 82. Review. A spool of wire of mass M and radius R is top is a tom of (c) Fin tal be for 85. A t S wit tal (a) bef No De bef 86. Re the as Example: Rolling, pg333, # 81 Energy conservation: ∆K + ∆U = 0 1 2 1 2 Iω + mv − 0 + (0 − mg (R − r )(1 − cos θ)) = 0 2 2 v = rω 1 2 2 2 1 ( mr )ω + m(r ω)2 2 5 2 + mg (R − r )(cos θ − 1) = 0 7mr 2 ω2 + 10mg (R − r )(cos θ − 1) = 0 r ω= 10g (R − r )(1 − cos θ) 7r 2 eter R and also of mass M. What is the kinetic energy of theAcam–shaft combination when Example: Trebuchet, pg328, # it47is rotating with angular speed v about the shaft’s axis? y 47. A war-wolf or trebuchet is a device used during the Mid- x 43 s conotates 00 m x .00 m 4.00 m Q/C dle Ages to throw rocks at castles and now sometimes used to fling large vegetables and pianos as a sport. A simple trebuchet is shown in Figure P10.47. Model it as a stiff rod of negligible mass, 3.00 m long, joining particles of mass m1 5 0.120 kg and m 2 5 60.0 kg at its ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 14.0 cm from the large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation. (a) Find the maximum speed that the small-mass object attains. (b) While the small-mass object is gaining speed, does it move with constant acceleration? (c) Does it move with constant tangential acceleration? (d) Does the trebuchet move with constant angular acceleration? (e) Does it have constant momentum? (f) Does the trebuchet–Earth system have constant mechanical energy? nstant momentum? (f) Does the trebuchet– Example: Trebuchet stem have A constant mechanical energy? m1 m2 3.00 m Figure P10.47 Example: A Trebuchet (a) Maximum speed of small mass? Example: A Trebuchet (a) Maximum speed of small mass? Will occur when arm is vertical, because up until that point gravity is providing a clockwise torque and doing positive work. (After, the work is negative) Example: A Trebuchet (a) Maximum speed of small mass? Will occur when arm is vertical, because up until that point gravity is providing a clockwise torque and doing positive work. (After, the work is negative) The rod has negligible mass. For the rest of the system, use energy conservation: ∆K + ∆U = 0 1 2 Iω − 0 + m1 g (3 − 0.14) + m2 g (0 − 0.14) = 0 2 Solve for ω, v = r ω. v = 24.5 m s−1 Summary • rolling motion • practice problems Next test Friday, June 5. (Uncollected) Homework Serway & Jewett, • From last time: Ch 10, page 328-. Questions: Section Qs 45, 47, 51, 53 • Ch 10, page 328-. Questions: Section Qs 49, 59, 65, 69
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