Notes on Torque. We ve seen that if we define torque as rfsinθ, and the N 2. i i

Transcription

1 Notes on Torque We ve seen that if we define torque as rfsinθ, and the moment of inertia as N, we end up with an equation mr i= 1 that looks just like Newton s Second Law There is a crucial difference, though: Mass is an intrinsic property of an object, or system of objects Moment of inertia, on the other hand, also depends on the axis about which we re rotating In other words, I really should have written the equation as: τ = I i i α z z z to emphasize that I was considering rotations about the z axis

2 Torque as a Vector The fact that I depends on which axis we re talking about means that one can t write an equation like this: τ = Iα But one can still define a vector form of torque: τ = r F this is consistent with the magnitude of the torque we ve defined One final note: unless mass and force, moment of inertia and torque depend on the frame of reference

3 Moment of Intertia for Solid Objects We have also seen that the moment of interia or rotational inertia I defines how hard it is to give a system an angular acceleration Just as mass defines how hard it is to give a system a linear acceleration For any collection of point particles, I is given by: I N = mi r i= 1 But in many cases we ll be dealing with solid objects, where mass is distributed over some volume. In this case, the sum becomes an integral: I = r dm i r is the perpendicular distance to the chosen axis

4 Example: Thin Rod We want to find the moment of inertial for a uniform thin rod of mass M about the axis shown: L/ L/ If we call the horizontal axis the x axis, we have: dm = M L L I = x dx = = = L L 3 L 3 1 L / 3 L / M Mx M ML L / L / dx 3

5 We can apply a similar procedure to find the moment of inertia of other common objects like hoops, cylinders, spheres, plates, etc. A tabulation of these is given in Fig of the text. I won t expect you to memorize these! Fig has a limitation, though it only shows axes that go through the center of mass In principal, there are an infinite number of other choices possible, which raises the specter of needed to perform an integral in each case Fortunately, we don t need to do this

6 The Parallel-Axis Theorem Consider a general solid object, rotating about an axis perpendicular to the page: Center of mass y x ( ) I = r dm = x + y dm ( ( ) ( ) ) Axis of rotation = x + x x + y + y y dm

7 ( ( ) ( ) ) I = x + x x + y + y y dm ( [ ] [ ] ) x y = x + + y + dm Note that x and y are measured in a frame with the center of mass at the origin ( ) ( ) ( ) = x + y dm+ x x + y y dm+ x + y dm The first term is: ( ) d ( ) x + y m = x + y dm = Mr While the last term is just the expression for moment of inertia about an axis passing through the center of mass, I

8 We still have to deal with the middle term: ( ) + = + x x y y dm x x dm y y dm Integrals like these should look familiar they re the x and y positions of the center of mass in the primed reference frame But by definition, in this reference frame the center of mass is at the origin Thus both these integrals are zero, so we re left with: I = I + Mr This is the parallel axis theorem Note that this implies that an axis through the center of mass gives the smallest moment of inertia (in a given direction)

9 Example: Object Rolling Down a Ramp Consider the angular acceleration of two circular objects with the same mass and radius rolling (without slipping) down a ramp: Uniform cylinder Hollow cylinder θ θ First we need to pick our axis One might choose the axis through the center of mass, but I ll pick instead the point of contact between the object and the ramp

10 The next step is to make a force diagram for the object: Center of mass mg R f θ Axis of rotation Note that the frictional force supplies no torque about the axis we ve chosen Thus the net torque is that due to gravity: τ = mgrsinθ

11 So the angular acceleration of the object is: τ τ α = = I I + mr 1 For the uniform cylinder, I is, while for the hollow cylinder it s mr mr So the answer is: τ mgrsinθ g sinθ αuniform = = = I 3 + mr mr 3R τ mgrsinθ g sinθ 3 αhollow = = = = α I + mr mr R 4 Parallel axis theorem uniform

12 If the ramp is of length d, how long does is take each object to roll down? Since there is no slipping, the total angle through which the object rotates is φ = d/r. Using the equations for rotational kinematics with constant acceleration, we find: 1 d φ = φo + ωot+ αt = R 1 d αt = R t t t = uniform hollow d αr d 3R 3d = = R gsinθ gsinθ d R 4d = = R gsinθ gsinθ The answer doesn t depend on R!