Rotation Work and Power of Rotation Rolling Motion Examples and Review

Rotation Work and Power of Rotation Rolling Motion Examples and Review Lana Sheridan De Anza College Nov 22, 2017

Last time applications of moments of inertia Atwood machine with massive pulley kinetic energy of rotation

Warm Up Question: Kinetic Energy of Rotation Quick Quiz 10.6 1 A section of hollow pipe and a solid cylinder have the same radius, mass, and length. They both rotate about their long central axes with the same angular speed. Which object has the higher rotational kinetic energy? (A) The hollow pipe does. (B) The solid cylinder does. (C) They have the same rotational kinetic energy. (D) It is impossible to determine.

Overview work and power of rotation Rolling motion Examples

Work of Rotation We can define the work done by a torque τ over a rotation of angular displacement θ = θ f θ i : W = θf θ i τ dθ This is equivalent to the force integral definition.

Work of Rotation Two arguments to show they are the same. First: W = F ds

Work of Rotation Two arguments to show they are the same. First: W = F ds = F t ds = F t r dθ (ds = rdθ) = τ dθ (τ = r F t ) = τ dθ The last line follow because the τ and θ vectors point along the same (fixed) axis.

Work of Rotation Second argument, using the vector identity A (B C) = B (C A): W = F ds = F (dθ r) = dθ (r F) = τ dθ

Power W = τdθ implies: dw dθ = τ

Power W = τdθ implies: dw dθ = τ dw dt dt dθ = τ dw = τ dθ dt dt

Power W = τdθ implies: dw dθ = τ dw dt dt dθ = τ dw = τ dθ dt dt Giving an expression for power: P = τω

Work-Kinetic Energy Theorem W = = τdθ (Iα)dθ

Work-Kinetic Energy Theorem W = = = = = τdθ (Iα)dθ I dω dt dθ ( dω I dθ Iω dω dθ dt ) dθ

Work-Kinetic Energy Theorem W = τdθ = (Iα)dθ = I dω dt dθ ( ) dω dθ = I dθ dθ dt = Iω dω = ( ) 1 2 Iω2 W = K

Energy In total then, W = K trans + K rot + U + E int where K trans represents the kinetic energy of the CM motion, and K rot is the rotational kinetic energy

Rolling Motion translation Combination of translation and rotation A combination of translation and rotation. v CM v v CM R v 2v CM M v CM CM v v CM P v CM v 0 P c The CM is translated at velocity v CM. Figure 10 rolling obj a combina and pure r Notice that instantaneously P is the pivot point of the rotation.

Henry Leap and Jim Rolling Motion The pivot point of the rotation changes as different parts of the wheel contact the surface. anslational speed of (10.28) olds whenever a cylure rolling motion. ass for pure rolling R u s (10.29) t speed v CM, staying ass of the object. As s Ru Figure 10.24 For pure rolling motion, as the cylinder rotates through an angle u its center v CM = ds moves a linear distance dt = Rω s 5 Ru.

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 2 I Pω 2 Using the parallel axis theorem: I P = I CM + mr 2. K = 1 2 (I CM + mr 2 )ω 2 = 1 2 I CMω 2 + 1 2 mv 2 CM = K CM,rot + K CM,trans

Rolling down an incline Equation 10.30 to obtain A sphere starts from rest at the top of an incline and rolls down. Find the (translational) velocity of the center of mass at the Total kinetic energy bottom of the incline. of a rolling object R M x u h v S v CM Figure 10.26 A sphere rolling down an incline. Mechanical energy of the sphere Earth system is conserved if no slipping occurs. K 5 Using v CM 5 Rv, this equation ca The term 1 2I CM v 2 represents the center of mass, and the term 1 2Mv 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 K

Rolling down an incline Equation 10.30 to obtain A sphere starts from rest at the top of an incline and rolls down. Find the (translational) velocity of the center of mass at the Total kinetic energy bottom of the incline. of a rolling object R M x u h v S v CM Figure 10.26 A sphere rolling down an incline. Mechanical energy of the sphere Earth system is conserved if no slipping occurs. K 5 Using v CM 5 Rv, this equation ca The term 1 2I CM v 2 represents the center of mass, and the term 1 2Mv 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 K + U = 0 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 K

Rolling down an incline Equation 10.30 to obtain A sphere starts from rest at the top of an incline and rolls down. Find the (translational) velocity of the center of mass at the Total kinetic energy bottom of the incline. of a rolling object R M x u h v S v CM Figure 10.26 A sphere rolling down an incline. Mechanical energy of the sphere Earth ) system is conserved if no slipping occurs. ( 1 2 I CMω 2 + 1 2 Mv CM 2 0 K 5 Using v CM 5 Rv, this equation ca The term 1 2I CM v 2 represents the center of mass, and the term 1 2Mv 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 K + U = 0 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 + (0 Mgh) = 0 K

Rolling down an incline Equation 10.30 to obtain A sphere starts from rest at the top of an incline and rolls down. Find the (translational) velocity of the center of mass at the Total kinetic energy bottom of the incline. of a rolling object R M x u h v S v CM Figure 10.26 A sphere rolling down an incline. Mechanical energy of the sphere Earth ) system is conserved if no slipping occurs. ( 1 2 I CMω 2 + 1 2 Mv CM 2 0 v CM = K 5 Using v CM 5 Rv, this equation ca The term 1 2I CM v 2 represents the center of mass, and the term 1 2Mv 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 K + U = 0 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 + (0 Mgh) = 0 2gh 1 + (I CM /MR 2 )) K

Question Quick Quiz 10.7 2 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.

Figure 10.27 (Example 10.14) h as a marble and a croquet ball. here, a solid cylinder, or a hoop, we would obtain Example 10.14 - Pulling a Spool er. The constant factors that appear in the exprest the Acenter cylindrically of mass symmetric for the spool specific of mass object. m and In radius all R sits at rest lue the on acceleration a horizontal table would withhave friction. if the You incline pull on were on a light string 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. n a horizontal apped around l force of maga distance L R r L T S pool.

s than g sin u, the value the acceleration would have if the incline were Example 10.14 - Pulling a Spool M dius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of maglls without slipping a distance L R r L T S ter of mass of the spool. Can use: W = K. e motion of the spool when you istance L, notice that your hand nt 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 work energy relationships for introductory physics, The Physics Teacher, 43:10, 2005.

MExample 10.14 - Pulling a Spool dius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of maglls without slipping a distance L R r L T S ter of mass of the spool. Can use: W = K. e motion of the spool when you istance L, notice that your hand nt 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. W = TL This can be thought of in two ways. 1 + r R work energy relationships for introductory physics, The Physics Teacher, 43:10, 2005. continued First, the string unrolls from the spool, so the point of application of the force F (the hand) moves a distance L + L r R. Alternatively, there is work done translating the spool: W = TL, plus work done rotating the spool: W = τ θ = (rt ) ( r R ).

s than g sin u, the value the acceleration would have if the incline were Example 10.14 - Pulling a Spool M dius R sits at rest on a horizontal on a light string wrapped around constant horizontal force of maglls without slipping a distance L R r L T S ter of mass of the spool. Can use: W = K. e motion of the spool when you istance L, notice that your hand nt 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. ( W = TL 1 + r ) R work energy relationships for introductory physics, The Physics Teacher, 43:10, 2005. v CM = continued 2TL (1 + r/r) m + I/R 2

Example: Rolling, pg333, # 81 (b) Assuming the board is 1.00 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 a s

Example: Rolling, pg333, # 81 Energy conservation: ( K + U = 0 1 2 Iω2 + 1 ) 2 mv 2 0 + (0 mg(r r)(1 cos θ)) = 0 v = rω ( 1 2 (2 5 mr 2 )ω 2 + 1 ) 2 m(rω)2 + mg(r r)(cos θ 1) = 0 7mr 2 ω 2 + 10mg(R r)(cos θ 1) = 0 ω = 10g(R r)(1 cos θ) 7r 2

mine the moment of inertia of this structure about this axis. Example - Moment of Inertia and Rotational rotation KE Section 10.7 Rotational Kinetic Energy Page 328, #44 Axis of 45. The four particles in Figure P10.45 are connected by W rigid rods of negligible mass. The origin is at the cen- 44. Rigid rods of negligible mass lying along the y axis connect W three particles (Fig. P10.44). The system rotates Q/C about the x axis with an angular speed of 2.00 rad/s. y Find (a) the moment of inertia about the x axis, (b) the total rotational kinetic energy 4.00 kg y 3.00 m evaluated from 1 2Iv 2, (c) the x O tangential speed of each 2.00 kg y 2.00 m particle, and (d) the total kinetic energy evaluated from a 1 2m i v 2 i. (e) Compare the answers for kinetic energy in 3.00 kg y 4.00 m parts (a) and (b). Figure P10.44 x Figure P10.43 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? (b) & (d) Kinetic energy? I x = 92 kg m 2

Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? I x = 92 kg m 2 (b) & (d) Kinetic energy? K = 184 J (c) tangential speeds?

Example - Moment of Inertia and Rotational KE (a) Moment of inertia about x-axis? I x = 92 kg m 2 (b) & (d) Kinetic energy? K = 184 J (c) tangential speeds? v = rω

Summary rolling motion energy of rotation Collected Homework! Look for it later today. (Uncollected) Homework Serway & Jewett, PREV: Ch 10, onward from page 288. Probs: 45, 47, 49, 51, 53, 69 Ch 10, rolling motion problems, 59, 65. Conc. Question 13.