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Announcements

Announcements. Quiz #7 is this Friday: Unit 13: Impulse Unit 14: Rotational Kinematics You should also review energy conservation (or 1D kinematics with constant acceleration). Lecture 16: Rotational Dynamics. Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration.

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Announcements

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  1. Announcements Quiz #7 is this Friday: • Unit 13: Impulse • Unit 14: Rotational Kinematics • You should also review energy conservation (or 1D kinematics with constant acceleration)

  2. Lecture 16: Rotational Dynamics Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration

  3. I felt like every slide just had a ton of equations just being used to find new equations. Other than that the actual use of the equations doesn't seem so bad, but I still don't really know what to do for the CheckPoints. Energy Conservation H α F = ma f a Mg

  4. Let’s work through this again.

  5. Rolling Motion Objects of differentI rolling down an inclined plane: ΔK= -ΔU=Mgh v=0 ω=0 K=0 R M h v=ωR

  6. Rolling If there is no slipping: v ω v Wherev =ωR v In the lab reference frame In the CMreference frame

  7. Rolling c c c c Use v = ωRand I = cMR2 Hoop: c = 1 Disk: c = 1/2 Sphere: c = 2/5 etc… v So: Doesn’t depend on Mor R, just on c(the shape)

  8. ACT A hula-hoop rolls along the floor without slipping. What is the ratio of its rotational kinetic energy to its translational kinetic energy? c c A) B) C) Recall that I = MR2for a hoop about an axis through its CM: (c = 1 for a hoop)

  9. A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without frictionand the ball rolls without slipping. Which one makes it furthest up the ramp? CheckPoint: Block and Ball on Ramp A) Block B) Ball C) Both reach the same height. v ω v

  10. CheckPoint: Discuss and Revote The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp? A) Block B) Ball C) Same A) The ball initially converts some of it's potential energy into rotational KE and translational KE and therefore moves with less speed up the ramp, the block is only converting potential energy into translational KE. B) The ball has more kinetic energy to begin with because it has rotational and linear kinetic energy so it will get higher because it must end with more potential energy because of the law of conservation of energy. C) Energy conserved is the same..thus same height v ω v

  11. CheckPoint: Cylinder & Hoop on Ramp A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A) Cylinder B) Hoop C) Both reach the bottom at the same time

  12. CheckPoint: Discuss and Revote A) Cylinder B) Hoop C) Both reach the bottom at the same time A) The cylinder has a smaller moment of inertia so it accelerates faster. B) The hoop has a greater moment of interia so it will have greater rotational energy. C) They will have equal amounts of kinetic energy due to the fact they have the same mass and radius Hoop: c = 1 Cylinder: c = 1/2

  13. CheckPoint: Discuss and Revote A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first? A) Small cylinder B) Large cylinder C) Both reach the bottom at the same time A) The smaller cylinder has a smaller moment of inertia. 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 calculations, you go ahead and cancel out the masses and radii. That's a cancellation. Now you're left with an energy equation 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 time. Okay. Newton’s 2nd Law: For a cylinder, c = 0.5, so: Newton’s 2nd Law for rotations: So finally:

  14. What you saw in your Prelecture: Acceleration depends only on the shape, not on mass or radius.

  15. ACT A) Small cylinder B) Large cylinder C) Both reach the bottom at the same time A small light cylinder and a large heavy cylinder are released at the same time and slide down the ramp without friction. Which one reaches the bottom first? Newton’s 2nd Law: Newton’s 2nd Law for rotations: (no rotation)

  16. Moment of inertia of a cone: INTEGRATION!!! • The physics is in the setting up of the integral. • The calculus is in computing the integral. • You must understand the physical meaning of a differential volume: Cylindrical coords: Blue: z Red: ϕ Green: ρ Cartesian Cylindrical Spherical (polar)

  17. Moment of inertia of a cone: INTEGRATION!!! Total volume of cone: Cylindrical coords: Blue: z Red: ϕ Green: ρ …but ρ depends on z: The cone “looks” the same in all different directions, ϕ (azimuthal symmetry)

  18. ACT Suppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1 > m2) are attached 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 acceleration of the two masses: A) a1 > a2 B) a1 = a2 C) a1< a2 M ω R T2 T1 a2 m2 m1 a1

  19. ACT Suppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1 > m2) are attached 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. How is the angular acceleration of the wheelrelated to the linear acceleration of the masses? A) α = Ra B) α = a/R C) α = R/a M ω R T2 T1 a2 m2 m1 a1

  20. ACT Suppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1 > m2) are attached 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 tension in the string on either side of the pulley: A) T1 > T2 B) T1= T2 C) T1 < T2 M ω R T2 T1 a2 m2 m1 a1

  21. m2g m1g Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk-shaped pulley as shown. • Find the acceleration of the blocks. M ω R T2 T1 a m2 m1 a

  22. Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk-shaped pulley as shown. Find the acceleration of the blocks. For the pulley use τ=Iα m2g (Since for a disk) m1g • For the hanging masses useF = ma • m1 g -T1= m1a • -m2 g+T2=m2 a M ω R T2 T1 a m2 + + m1 T1R - T2R a

  23. We have three equations and three unknowns (T1, T2, a). Solve for a. m1 g -T1=m1a (1) -m2 g + T2= m2 a (2) T1-T2(3) m2 g m1g Atwood's Machine with Massive Pulley: M ω R T2 T1 a m2 + + m1 a

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