Chapter 12++ Revisit Circular Motion Revisit: Anular variables Second laws for radial and tanential acceleration Circular motion CM 2 nd aw with F net To-Do: Vertical circular motion in ravity Complete the rotatin rod story by combinin the force, enery conservation, and torque equations Tension within rotatin rods Vertical Circular Motion in Constant Gravity Rod Swinin Vertically on a Pivot We started the story of the fallin rod or stick pivoted on one end in Ch. 9++ and continued it in Ch. 11++ and we will complete it here. Note that the results also apply to a fallin trap door, because the moment of inertia for a thin door viewed on ede has the same form as that for a P thin rod. θ m ω, α I) Review: The radial and tanential components of the force on the rod due to the pivot are related by the translational second law to the anular velocity ω and the anular acceleration α (with the above picture, ω and α have been defined to be positive in the CW direction, for convenience). By now, we hope that the radial and tanential component picture (Ch. 12) is familiar to the reader: radial P tanential The radial and tanential components of the translational second law F net = M a CM have been found in the Ch. 11++ text and the problem solution, respectively: F P, radial msinθ = mω 2 / 2 A (p. 11-17) and F P, tanential + mcosθ = + mα / 2 Problem 11-8a (continued) 12-18
In summary, if we knew the anular velocity ω and the anular acceleration α, for a iven θ, the force vector due to the pivot on the rod would be completely determined by these equations. And by the way, in the desin of doors and brides and ladders, etc., we need to know these forces for personal and structural safety. The deree to which materials can withstand and bear up under weihts and other forces is vital to enineerin desin. This is the alpha and omea of our physics study! II) Review: The anular velocity ω(θ) can be predicted if we know an earlier anular velocity ω 0 (θ 0 ) from the conservation of enery. In Ch. 9++, we used conservation of enery to show that ω 2 = ω 2 0 + 3 (sin θ sin θ) B 0 We could now insert this expression into F P, radial msinθ = mω 2 / 2, but there s no lovely simplification in doin so, so let s wait to look at special cases. III) The missin piece: the anular acceleration α from the rotational second law. What remains is to show how to find α at a iven θ throuh τ net,p = I P α, the rotational second law form for a rotation of the rod around its end pivot point P. The first inredient is I P = 1 3 m2 for the rod around the axis perpendicular to and throuh its end (which we already used in Ch. 9++ to calculate the rotational kinetic enery). The torque due to the rod s weiht is + m cosθ because 1) we now use the convention CW is positive, 2) the CM 2 is at /2, and 3) F = mcosθ (or, if you prefer, you can use the equivalent torque formula, FRsin( π 2 + θ) = m cos θ, which ives the same answer). The 2 rotational second law, τ net,p = I P α, now reads + m 2 cosθ = + 1 3 m2 α or, simplifyin and rearranin, α = 3 2 cosθ Insertin this into F P, tanential + mcos θ = + mα / 2, we have C F P, tanential = mcosθ + m 3 2 ( cosθ) / 2, or F P, tanential = 1 4 mcos θ D (example on next pae) 12-19
Example: Suppose a thin uniform rod of mass m and lenth has its frictionless hine pivoted on a horizontal axis. It is lifted up to the anle of 30 o with respect to the horizontal plane and is released from rest, rotatin downward due to ravity. As a result, it swins down throuh θ = 0 havin some nonzero anular velocity at that horizontal position. Notice from our eneral formula C obtained from the rotational second law that and α 0 = 3 3 4 at the initial 30o position α = 3 at the horizontal position. 2 ω 0 = 0, α 0 ω, α What are the forces due to the pivot on the rod at the horizontal position (θ=0), after fallin from rest at the initial anle of 30 0? 1) First, let s o after F P, radial where we need ω 2. Note from the fiure on the riht, this is the horizontal force on the rod due to the pivot. P tanential is vertical now radial is horizontal now With ω 0 = 0 at θ 0 = 30 o, the enery result B, ω 2 = ω 2 0 + 3 (sin θ sin θ), reduces 0 to: ω 2 = 3 (0 sin30o ) = + 3 2 So formula A, F P, radial - msinθ = m ω 2 / 2 turns into F P, radial 0 = m 3 2 / 2, or F P, radial = 3 4 m at θ = 0 (this is only true for the fall from rest at 300 ) Indeed, the pivot has to pull back to the left on the rod as it swins down (to make it o in a circle!) This component, - ¾m, is horizontal and points to the left, like a ood centripetal force should for the rod in a horizontal position. (continuin next to the tanential force component) 12-20
2) Now for F P, tanential, which, at θ = 0, is a vertical component (see the previous fiure). Pluin into formula D, F P, tanential = 1 mcos θ, we et 4 F P, tanential = 1 m at θ = 0 (but this is independent of the initial conditions) 4 From the minus sin (down was positive), the pivot thus pushes vertically up on the rod in reaction to its weiht, at θ = 0. But notice it is NOT 1 m as it would 2 be if the rod were supported on both ends and wasn t rotatin down. The interestin fact is that the pivot supports less than half the weiht when the rod is swinin down!! Problem 12-8 a) To feel better about bein able to derive thins from the basics without just pluin into our eneral formulas, derive the instantaneous anular acceleration 3 result α = of a rod, just as it passes throuh the horizontal position, 2 startin with the rotational second law at that position. 3 b) If α = is the instantaneous anular acceleration of a rod just as it 2 passes throuh the horizontal position, then what is the linear acceleration downward (i.e., the tanential acceleration) of a point on the rod a distance r from the pivot? c) Consider the followin values r =, /2, and 2/3 in your formula from (b). Discuss anythin interestin you find. There is a neat demonstration related to this problem utilizin pennies on a meter stick see your lecture! Problem 12-9 A thin rod of uniform mass m and lenth is released from rest at the horizontal position, as shown, with its left end pivoted at a frictionless hine attached to a ceilin corner. It falls due to ravity but is constrained by the pivot as it falls. a) Usin formulas A, B, and D, of pp. 12-18 and 12-19, find, as the rod passes throuh the downward vertical position, the vertical (F P,radial F V ) and horizontal (F P,tanential F H ) components of the force exerted on the rod by the pivot in terms of m,, or nothin!,, also shown. Initial: Final: b) As a separate exercise, derive F V and F H from first principles: namely, the translational second law, the rotational second law, and conservation of enery. 12-21
Vertical Circular Motion in Constant Gravity We re oin on a diet riht now and we re shrinkin all our masses down to points so we won t need any moments of inertia in this section. We instead want to concentrate on combinin enery and centripetal motion. Example: A small mass m is bein swun in a vertical circle at the end of a massless rod of lenth. The other end of the rod is connected to a frictionless pivot. Suppose the speed of the mass at the top of the circle is v. What is its speed u(θ) when the rod makes an anle θ with the vertical? Beinnin: ater: Answer: A statement about the frictionless pivot is a clue to use conservation of enery. The pivot force does no work. Thus K + U at the top is equal to K + U anywhere else (in particular, at anle θ) where U = U(constant ravity) = my: K i + U i = 1 2 m v2 + m y top K f + U f = 1 2 m u2 + m y(θ) 1 2 m v2 + m y top = 1 2 m u2 + m y(θ) Set the oriin y=0 at the pivot (wherever you choose your oriin to be, it wouldn t matter the difference in y value would cancel out on both sides and you d et the same answer). So y top =, and y(θ) = cosθ from a olden trianle. With the mass m cancelin out, atherin terms and multiplyin by 2, we have the answer for the speed u(θ): u 2 = v 2 + 2 (1 - cosθ) After a quick diression to remind you of the equivalent, and perhaps your preferred way of writin enery conservation, we pose another question about the rod s force on m as a function of θ - which we can answer now with the help of the expression for u(θ). 12-22
Diression: Recall the old story about how we may replace K i + U i = K f + U f by ΔK = - ΔU, the equivalent form of enery conservation. (Δ means the chane made in oin from the initial state to the final state). We first say ΔK = ΔU = m Δy Then we use Δy = y f y i = cos θ plus ΔK = 1 2 m u2 1 2 m v 2 and we recover the previous result. It is really the same calculation but whatever helps you make the fewest errors, o ahead and use it. As we said, we want to answer the followin question: How does the manitude and sin of the radial force on m due to the rod in the previous example chane as a function of θ as the rod swins around? Answer: Use the second law alon the radial direction to et an equation for the outward radial force N on m due to the rod. (Callin the outward axis positive, the radial component of the force on m due to the rod is defined as +N. If we find N < 0, then this means the rod is pullin inward on m). The circular motion means the radial acceleration is centripetal and inward (so its component is neative). For the previous convention (velocity u at the anle θ), we find, from the FBD shown, the second law alon the radial direction is N m cosθ = m u2 with a olden trianle calculation yieldin -mcosθ, the radial component, for the weiht. θ m Also, since the centripetal acceleration involves u 2, we can use the enery conservation result for u 2 at θ, and relate it to the oriinal velocity v: Solvin for N: or, finally, N m cosθ = m v 2 N = m cosθ m v 2 (comments on the next pae) + 2 (1 - cosθ) 2 m (1 - cosθ) N = m (3cosθ 2) m v 2 12-23
Comments: If N > 0, we have compression (the rod pushes up on m) If N < 0, we have tension (the rod pulls in on m) Indeed, with N = m (3cosθ 2) mv 2 /, we et the expected result that the rod will have to pull inward on the mass if it is oin really fast (e.., N < 0 in the above expression for lare v). And, for small θ (i.e., near the top), we et the expected result that the rod would have to push out to hold the mass up, if it s hardly movin (e.., N > 0 in the above for smaller v and smaller θ. Problem 12-10 A small mass m attached by a very liht metal rod of lenth to a frictionless pivot moves under constant ravity in a vertical circle. At the bottom of the circle suppose it has a speed iven by v 0 = 2 ( ) (this is a contrived choice in terms of the iven parameters in order to simplify the problem calculation!!) a) What is and is not conserved in this problem? Why and why not? b) What is the speed v of m when it is 30 above the horizontal, as shown, in terms of and? c) What is the tension or compression (which is it?) in the rod when m is in the 30 position, in terms of m and? d) In eneral, is a rod swinin around like this always under tension, or can it be under compression at some point as it swins around? What does the question about tension and compression depend upon? (Note that it would always be under tension at the bottom why?) 12-24