* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Newton`s Second Law for Rotation Newton`s First Law for Rotation
Classical mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Belt (mechanical) wikipedia , lookup
Coriolis force wikipedia , lookup
Fictitious force wikipedia , lookup
Work (physics) wikipedia , lookup
Classical central-force problem wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Rotational spectroscopy wikipedia , lookup
Equations of motion wikipedia , lookup
Jerk (physics) wikipedia , lookup
Continuously variable transmission wikipedia , lookup
Center of mass wikipedia , lookup
Moment of inertia wikipedia , lookup
Proper acceleration wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Centripetal force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
PY105 (C1) • Assignment 9 has been posted on WebAssign. • The grades for Test 2 have been posted on WebAssign. Newton’s Second Law for Rotation The equation, v v ∑ τ = Iα v v is the rotational equivalent of ∑ F = ma. Torque plays the role of force. Rotational inertia plays the role of mass. Angular acceleration plays the role of the acceleration. Newton’s First Law for Rotation An object at rest tends to remain at rest, and an object that is spinning tends to spin with a constant angular velocity, unless it is acted on by a nonzero net torque or there is a change in the way the object's mass is distributed. Rotational Inertia How hard it is to get something to spin, or to change an object's rate of spin, depends on the mass, and on how the mass is distributed relative to the axis of rotation. Rotational inertia, or moment of inertia, accounts for all these factors. The moment of inertia, I, is the rotational equivalent of mass. The net torque is the vector sum of all the torques acting on an object. The tendency of an object to maintain its state of motion is known as inertia. For straight-line motion mass is the measure of inertia, but mass by itself is not enough to define rotational inertia. For an object like a ball on a string, where all the mass is the same distance away from the axis of rotation: I = mr 2 If the mass is distributed at different distances from the rotation axis, the moment of inertia can be hard to calculate. It's much easier to look up expressions for I from the table on page 291 in the book (page 10-15 in Essential Physics). The parallel axis theorem A table of rotational inertias If you know the rotational inertia of an object of mass m when it rotates about an axis that passes through its center of mass, the object’s rotational inertia when it rotates about a parallel axis a distance h away is: I = ICM + mh 2 Take a ring with radius R and mass M as an example: Axis of rotation x I = MR2 x I = MR2 + MR2 = 2MR2 1 Example 1. Torque on a pulley Example 1. Torque on a pulley (cont’d) A constant force of F = 8 N is applied to a string wrapped around the outside of the pulley. The pulley is a solid disk of mass M = 2.0 kg and radius R = 0.50 m, and is mounted on a horizontal frictionless axle. What is the pulley's angular acceleration? Solution: Simulation v v ∑ τ = Iα RF = 1 MR 2 2 1 MRα 2 α= 2 × ( 8.0 N) 2F = = 16 rad/s2 MR (2.0 kg) × (0.5 m) R F Example 2. Two identical pulleys (cont’d) Example 2. Two identical pulleys Simulation We take two identical pulleys, both with string wrapped around them. On the one on the left we apply an 8 N force to the string. On the one on the right we hang an object with a weight of 8 N. Which pulley has the larger angular acceleration? α1 α2 1. The one on the left R R 2. The one on the right 3. I= F= Why are we told that the pulley is a solid disk? So we know what to use for the rotational inertia. I = 1 MR 2α 2 α MR2/2 Neither, they're equal F = 8N For the pulley on the left, the tension in the string is 8 N. For the system on the right, let’s draw the free-body diagram of the weight. α2 T τ = RT R T a = (8N-T)/m mg = 8N What does the free-body diagram tell us about the tension in the string? For the weight to have a net force directed down, the tension must be less than the force of gravity. So, the tension is less than 8 N. mg = 8N Atwood’s machine Atwood’s machine involves one pulley, and two objects connected by a string that passes over the pulley. In general, the two objects have different masses. We have encountered this in one of the homework assignments. a Re-analyzing the Atwood’s machine When we analyzed Atwood’s machine earlier, we found an expression for the acceleration of the weights in terms of m, M, and g, but we neglected the mass of the pulley (i.e., we assumed that the pulley is massless). If we include the mass of the pulley in the analysis, the acceleration of the masses m and M should be: 1. larger than before. 2. smaller than before. 3. the same as before. 2 Example 3 Acceleration in an Atwood’s machine Find an expression for the acceleration of m and M in an Atwood’s machine given the mass of the pulley is mp. Solution: What’s the difference with the earlier case where the pulley is massless? The acceleration turns out to be less than what we found before, because we need to accelerate the pulley – a solid disc (with I = ½ mpR2, where R is the radius of the pulley). Example 3 Acceleration in an Atwood’s machine (cont’d) Step 1: Analyze the lighter object Sketch a free-body diagram for the lighter object. Choose a positive direction, and apply Newton’s Second Law. Let’s choose positive to be up, in the direction of the acceleration. FT1 FT1 FT1 Because of this, the tension force on the right (where the heavier mass M is) is larger than the tension on the left to give a net clockwise torque to produce a clockwise angular acceleration to the pulley. Example 3 Acceleration in an Atwood’s machine (cont’d) Step 2: Analyze the heavier object Sketch a free-body diagram for the heavier object. Choose a positive direction, and apply Newton’s Second Law. Choose positive down this time, to match the object’s acceleration. FT2 v v ∑ F = Ma a +Mg − FT 2 = +Ma v v ∑ F = ma FT2 a FT2 +FT 1 − mg = +ma mg Example 3 Acceleration in an Atwood’s machine (cont’d) Step 3: Analyze the pulley Sketch a free-body diagram for the pulley. Choose a positive direction, and apply Newton’s Second Law for rotation. Choose positive clockwise, to match the pulley’s angular acceleration. v v ∑ τ = Iα α ⎛1 ⎞ +RFT 2 − RFT 1 = + ⎜ mpR 2 ⎟ α I = ½mpR2 ⎝2 ⎠ FT1 FT2 ⎛1 ⎞a +FT 2 − FT 1 = + ⎜ mpR ⎟ ⎝2 ⎠R 1 +FT 2 − FT 1 = + mpa 2 (Note that R cancels out so it is not needed to be specified.) Mg Example 3 Acceleration in an Atwood’s machine (cont’d) Step 4: combine the equations Lighter object: +FT 1 − mg = + ma Heavier object: +Mg − FT 2 = +Ma Pulley: 1 +FT 2 − FT 1 = + mpa 2 Add the equations: +Mg − mg = +Ma + ma + Previous result 1 ⎛ ⎞ +Mg − mg = ⎜ M + m + mp ⎟ a 2 ⎝ ⎠ a= Mg − mg M +m a= 1 mp a 2 Mg − mg m M +m+ p 2 3