1 / 14

Rotational Dynamics

Rotational Dynamics. The action of forces and torques on rigid object: Which object would be best to put a screw into a very dense, hard wood? A B= either C. Torque = Amount of Force x Lever Arm T = F L. Nm = N x m.

Download Presentation

Rotational Dynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Rotational Dynamics The action of forces and torques on rigid object: Which object would be best to put a screw into a very dense, hard wood? A B= either C

  2. Torque = Amount of Force x Lever ArmT = F L Nm = N x m If a child pulls the end of the end of the wrench with 10 N of force and an adult pulls at the ½ point of the wrench with 15 N, who pulls with the most torque and by how much more? ( ) = child = adult The child provides (.25) Nm more Torque Child T = 10N (.1) = 1 Nm Adult T = 15N (.05) = .75 Nm .1 m

  3. Torque If a person uses a wrench with 40 Nm of torque to put in a screw, how much force would be needed using a screwdriver to put n the screw? T = F L 40 Nm = F .01 (the lever arm of the screwdriver – ½ diameter) F = 4,000 N (4000N x .22 N/lb = 880 lbs of force) (Compare this with 88 lbs of force with the wrench – 40Nm/.1=F) .02 m .1 m

  4. Rigid Objects in Equilibrium For a rigid object to be in equilibrium, the objects linear motion and angular motion must not be accelerating. To identify if a rigid object is in equilibrium: Find all the external forces acting on an object, if all the forces along the x and y axis = 0 Fx= 0 Fy= 0, the net torque on the object should also = 0 T = 0

  5. Ignoring the mass of the board, how long far away from the fulcrum must Johan stand to balance the car? To balance the system, the torque on each end must be equal. F L = F L 490N (L) = 4,900 (3) L = 30m 50 kg 500 kg ? m 3 meters

  6. If the board has a mass of 500 kg (evenly distributed, where is the center of mass of the system? Center of Mass = m1r1 + m2r2 + m3r3 m1+m2+m3 50 kg 500 kg 1m 2m 30m

  7. 9.4 Moment of Inertia is the measure of mass distribution about an object rotating about an axis. I = mr2 (this can change due to the shape of the object) see page244 in your book What is I for a ring that has a mass of .01 kg and a diameter of .02m? What is I for a solid disc that has a mass of .01 kg and a diameter of .02m? Consider a hollowed out ball (A) (like a tennis ball) and a solid ball (B)(like a baseball) that have the same mass and radius, rolling across a floor, which comes to a stop first? (C) = come to a stop at the same time

  8. Rotational Motion about a fixed axis F = ma (linear) T = Iα (rotational) torque = Moment of Inertia x angular acceleration (needs to measured in radians) 360o = 2π radians How many radians are in 180o? Angular acceleration = ωf2 – ωo2ωf - ωo α 2Ɵ (in radians) t ω=angular velocity ω = v/r A circular saw accelerates from rest to 80 rev/sec in 240 revolutions, the blade has a mass of 1.5 kg and a radius of .2 meters. What torque must be applied to the motor to accelerate the blade? T = Iα I = α =

  9. 3 bars that pivot at the red point have weights on them with 3 units of weight distributed at different points along the bar. A B C A constant force is applied upward to the end of each bar. Rank which one accelerates the most to least. A = the most B = 2nd most C = the least

  10. Work = Fd (linear) Wrotational = TƟ(in radians) How much work is done by a pulley that produces 10 Nm of torque for one revolution?

  11. KE = ½ mv2 (linear) (rotational) KErotational = ½ Iω2 What is the KE of a hollow ring with mass 2kg and .25m radius rotating at 20 rad/sec.? Total Energy = Translational KE + Rotational KE +mgh E = ½ mv2 + ½ Iω2 + mgh The ring mentioned above rolls down a ramp meters high, how fast is the ring going at the bottom? ½ mv2 + ½ Iω2 = mgh ½ m v2 + ½ mr2 (v/r) = mgh 2mgh V = m + I/r2 5m

  12. Momentum Linear = mv Angular Momentum L = I ω For an object moving with a constant angular velocity around an axis, the momentum is conserved should the velocity or moment of inertia change. Why is it important for a diver who wants to do many flips before hitting the water to get in a tuck position? A disc (bottom) spins at a constant rate. A smaller disc is dropped on the larger disc. How does the new object (small and large together) behave? What if you dropped the hollow ring instead of the smaller disc?

  13. Torque = F l (lever Arm) For an object to be balanced, the torques on each side of the center of mass must be equal. Rigid objects in equilibrium – All forces are 0, not translational accel, no angular accel, no torque Center of Gravity – The of mass of a rigid object is the point where its entire weight can be considered. Newton’s 2nd Law (F=ma) for rotating objects: T = I α where t= torque, I=moment of inertia, α angular accel Moment of inertial = distribution of mass of a rotating object. I = mr2 (this can change based on shape of object) Rotational Work = TƟ, like W=Fd, T is torque, Ɵ is distance around the axis (in radians) Rotational Kinetic Energy, KEr = ½ Iω2, angular velocity Total Energy of an object = ½ mv2 + ½ Iω2 + mgh Angular Momentum, L = Iω

More Related