Rolling Rolling Condition for Rolling Without Slipping, Lecture notes of Physics

Download Rolling Rolling Condition for Rolling Without Slipping and more Lecture notes Physics in PDF only on Docsity! 1 1 Rolling 2 Rolling Rolling simulation We can view rolling motion as a superposition of pure rotation and pure translation. For rolling without slipping, the net instantaneous velocity at the bottom of the wheel is zero. To achieve this condition, 0 = vnet = translational velocity + tangential velocity due to rotation. In other wards, v – rω = 0. When v = rω (i.e., rolling without slipping applies), the tangential velocity at the top of the wheel is twice the translational velocity of the wheel (= v + rω = 2v). +rω -rω ω v +rω -rω ω v+ = Pure rotation Pure translation Rolling v v 3 Condition for Rolling Without Slipping When a disc is rolling without slipping, the bottom of the wheel is always at rest instantaneously. This leads to ω = v/r and α = a/r where v is the translational velocity and a is acceleration of the center of mass of the disc. v rω -rω ω X Vnet at this point = v – rω 4 Big yo-yo A large yo-yo stands on a table. A rope wrapped around the yo-yo's axle, which has a radius that’s half that of the yo-yo, is pulled horizontally to the right, with the rope coming off the yo-yo above the axle. In which direction does the yo-yo move? There is friction between the table and the yo-yo. Suppose the yo-yo is pulled slowly enough that the yo-yo does not slip on the table as it rolls. 1. to the right 2. to the left 3. it won't move 5 Big yo-yo Since the yo-yo rolls without slipping, the center of mass velocity of the yo-yo must satisfy, vcm = rω. With this, the direction of vcm (= direction of motion) is determined entirely by whether the yo-yo rolls clockwise or counter- clockwise. In this example, the tension on the rope produces a clockwise torque, which would cause the yo-yo to roll clockwise. So the yo-yo will move to the right. 6 Big yo-yo, again The situation is repeated but with the rope coming off the yo-yo below the axle. If the rope is pulled to the right, which way will the yo-yo move now? Again, suppose that there is no slip when the yo-yo rolls. 1. to the right 2. to the left 3. it won't move 2 7 Analyzing the yo-yo The key to determining which way the yo-yo moves is to look at the torque due to the tension about the point of contact. All the other forces acting at the point of contact will contribute no torque about the point of contact. α a To realize this motion, the net torque about the center of the yo-yo (= the torque due to static friction, τFs, plus the torque due to FT, τFT) must be clockwise. Since τFT is counterclockwise, this means that τFs must be clockwise and bigger than τFT. This requires FS to be pointing left and bigger than FT/2 (so that FSR > FTR/2). At the same time, FT must be bigger than FS in order to produce an acceleration that’s pointing right. Altogether, FT > FS > FT/2. 8 Big yo-yo Given the axle of a yo-yo to be half the radius of the yo- yo, and the yo-yo moves a distance L to the right when the rope is pulled from above the axle, how far does the end of the rope move? Again, the yo-yo rolls without slipping. 1. it doesn't move at all 2. L/3 6. 4L/3 3. L/2 7. 3L/2 4. 2L/3 8. 5L/3 5. L 9. 2L ω 9 The distance moved by the rope Given the axle to be half the yo-yo's radius, a point on the outer edge of the axle has a rotational (tangential) speed equal to half the yo-yo's translational speed. Let’s call the translational speed v. Above the axle, where the rope is unwinding, the net velocity is 1.5 v (= v + ½ rω = v + v/2). If the yo-yo moves a distance L, the end of the rope would move a distance 1.5 L. ω v 10 An accelerating cylinder A cylinder of mass M and radius R has a string wrapped around it, with the string coming off the cylinder above the cylinder. If the string is pulled to the right with a force F, what is the acceleration of the cylinder if the cylinder rolls without slipping? (Hint: Consider the direction of the frictional force and whether it will assist or resist F.) =1. Fa m <2. Fa m >3. Fa m F α 11 We would expect the frictional force to be pointing forward since the tensional force would produce a torque that rotates the cylinder clockwise, which produces a tendency for the bottom of the cylinder to move backward relative to the ground. Hence, the net force would be the sum of the frictional force and the tensional force (bigger than the tensional force along). Notice that the friction is static friction since there’s no slip, which would mean that the bottom of the wheel is momentarily at rest relative to ground. F Fs α Simulation An accelerating cylinder 12 An accelerating cylinder – Finding a and Fs A cylinder of mass M and radius R has a string wrapped around it, with the string coming off the cylinder above the cylinder. If the string is pulled to the right with a force F, what is the acceleration of the cylinder if the cylinder rolls without slipping? What is the frictional force acting on the cylinder? F α