Lecture 16: Rotational Dynamics
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- Gervase Parrish
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1 Lecture 6: otational Dynamics Today s Concepts: a) olling Kine6c Energy b) Angular Accelera6on Mechanics Lecture 6, Slide
2 I felt like every slide just had a ton of equa6ons just being used to find new equa6ons. Other than that the actual use of the equa6ons doesn't seem so bad, but I s6ll don't really know what to do for the CheckPoints. H PE KE = = MgH 0 Energy Conserva6on PE = 0 KE = K trans + K rot f α a F = ma Mg Mechanics Lecture 6, Slide
3 Let s work through this again. Mechanics Lecture 6, Slide 3
4 CheckPoint: Discuss and evote A small light cylinder and a large heavy cylinder are released at the same 6me and roll down a ramp without slipping. Which one reaches the bovom first? A C B D A) Small cylinder B) Large cylinder C) Both reach the bovom at the same 6me A) The smaller cylinder has a smaller moment of iner6a. B) Because the larger cylinder has more mass further from the center of mass. C) Let's break this down Hubie Brown- style. We have two cylinders. We are rolling these cylinders down a grey painted area. Your cylinders have different masses and different radii. Now, if I'm a student, I'm asking myself, Do I care about these differences? And the answer is no, okay. We do not care. Why don't we care? Well, because when you perform the calcula6ons, you go ahead and cancel out the masses and radii. That's a cancella6on. Now you're leu with an energy equa6on in which all the values are equal, and you have a high percentage chance to score a correct answer, okay. A correct answer, which is that both of these cylinders will reach the end of the painted area at the same 6me. Okay. Mechanics Lecture 6, Slide 4
5 olling Without Slipping Objects of different I rolling down an inclined plane: h M v = 0 ω = 0 K = 0 Δ K = W net = -ΔU = M gh K = Iω + Mv v = ω Mechanics Lecture 6, Slide 5
6 olling Without Slipping If there is no slipping: v v ω v Where v = ω In the lab reference frame In the CM reference frame Mechanics Lecture 6, Slide 6
7 olling Without Slipping K = Iω + Mv Use v = ω and I = cm v K = M ω + Mv = + c ( c ) Mv Hoop: c = Disk: c = / So: ( c + ) Mv = Mgh Sphere: c = /5 etc v = gh c + Doesn t depend on M or, just on c (the shape) Mechanics Lecture 6, Slide 7
8 CheckPoint: Discuss and evote A small light cylinder and a large heavy cylinder are released at the same 6me and roll down a ramp without slipping. Which one reaches the bovom first? A C B D A) Small cylinder B) Large cylinder C) Both reach the bovom at the same 6me A) The smaller cylinder has a smaller moment of iner6a. B) Because the larger cylinder has more mass further from the center of mass. C) Newton s Let's break this down Hubie Brown- style. We have two cylinders. We are rolling these cylinders down nd Law: For a cylinder, c = 0.5, so: a grey painted area. Your cylinders have different masses and different radii. Mgsin Now, if I'm fa s student, = MaI'm x asking myself, Do I care about these differences? And the I = answer is no, okay. We do not care. Why don't we care? Well, M because when you perform the calcula6ons, you go ahead and cancel out the masses and radii. That's a cancella6on. Newton s Now you're leu nd Law for with an energy equa6on in which all the values are equal, and you have a high percentage chance to score a correct answer, okay. A correct answer, which is that net = I = I a rotations: x =) f s = I a So finally: x a x = both of these cylinders will reach the end of the painted area at the same 6me. 3 g sin Okay. f s =) f s = Ma x Mechanics Lecture 6, Slide 8
9 ACT A small light cylinder and a large heavy cylinder are released at the same 6me and roll down a ramp without slipping. Which one is rota6ng faster at the bovom? A C B D A) Small cylinder B) Large cylinder C) Same If they reach the bottom at the same time, then they have the same CM speed at the bottom: V = V! =!! =! Mechanics Lecture 6, Slide 9
10 ACT A small light cylinder and a large heavy cylinder are released at the same 6me and roll down a ramp without slipping. Which one has more kine6c energy at the bovom? A) Small cylinder B) Large cylinder C) Same A C B D olling without slipping: c ( c ) Mv (c = 0.5 for a solid cylinder) K = M ω + Mv = + Mechanics Lecture 6, Slide 0
11 CheckPoint: Cylinder & Hoop on amp A cylinder and a hoop have the same mass and radius. They are released at the same 6me and roll down a ramp without slipping. Which one reaches the bovom first? A) Cylinder B) Hoop C) Both reach the bovom at the same 6me Mechanics Lecture 6, Slide
12 A) Cylinder B) Hoop CheckPoint: Discuss and evote C) Both reach the bovom at the same 6me A C B D A) The cylinder has a smaller moment of iner6a so it accelerates faster. B) The hoop has a greater moment of interia so it will have greater rota6onal energy. C) They will have equal amounts of kine6c energy due to the fact they have the same mass and radius v bottom = p r gh c + Hoop: c = Cylinder: c = / Mechanics Lecture 6, Slide
13 ACT A hula- hoop rolls along the floor without slipping. What is the ra6o of its rota6onal kine6c energy to its transla6onal kine6c energy? A C B D Krot A) = Krot 3 B) = C) Ktrans K 4 trans K K rot trans = ecall that I = M for a hoop about an axis through its CM: c ( c ) Mv (c = for a hoop) K = M ω + Mv = + Mechanics Lecture 6, Slide 3
14 A block and a ball have the same mass and move with the same ini6al velocity across a floor and then encounter iden6cal ramps. The block slides without fric6on and the ball rolls without slipping. Which one makes it furthest up the ramp? A) Block B) Ball CheckPoint: Block and Ball on amp C) Both reach the same height. v ω v Mechanics Lecture 6, Slide 4
15 CheckPoint: Discuss and evote The block slides without fric6on and the ball rolls without slipping. Which one makes it furthest up the ramp? v A) Block B) Ball C) Same ω v A C B D A) The ball ini6ally converts some of it's poten6al energy into rota6onal KE and transla6onal KE and therefore moves with less speed up the ramp, the block is only conver6ng poten6al energy into transla6onal KE. B) The ball has more kine6c energy to begin with because it has rota6onal and linear kine6c energy so it will get higher because it must end with more poten6al energy because of the law of conserva6on of energy. C) Energy conserved is the same..thus same height Mechanics Lecture 6, Slide 5
16 What you saw in your Prelecture: Acceleration depends only on the shape, not on mass or radius. Mechanics Lecture 6, Slide 6
17 ACT A C B D Suppose a cylinder (radius, mass M) is used as a pulley. Two masses (m > m ) are avached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the magnitudes of the accelera6on of the two masses: ω M A) a > a B) a = a C) a < a T T a m m a Mechanics Lecture 6, Slide 7
18 ACT Suppose a cylinder (radius, mass M) is used as a pulley. Two masses (m > m ) are avached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. A C B D How is the angular accelera6on of the wheel related to the linear accelera6on of the masses? ω M A) α = a B) α = a/ C) α = /a T T m m a a Mechanics Lecture 6, Slide 8
19 ACT Suppose a cylinder (radius, mass M) is used as a pulley. Two masses (m > m ) are avached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. A C B D Compare the tension in the string on either side of the pulley: A) T > T ω M B) T = T T T C) T < T a m m a Mechanics Lecture 6, Slide 9
20 Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk- shaped pulley as shown. Ø Find the accelera6on of the blocks. ω M T T a m m m g a m g Mechanics Lecture 6, Slide 0
21 Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk- shaped pulley as shown. Ø Find the accelera6on of the blocks. For the hanging masses use F = ma m g - T = m a -m g + T = m a ω M For the pulley use τ = Iα T - T (Since I = = I a M = = Ma I a for a disk) + T a T m m a m g m g + Mechanics Lecture 6, Slide
22 Atwood's Machine with Massive Pulley: We have three equa6ons and three unknowns (T, T, a). Solve for a. m g - T = m a () -m g + T = m a () ω M T - T (3) = Ma + T a T m m + a = m m + m m + M g m g m g a Mechanics Lecture 6, Slide
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