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Torque on a Pulley

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Torque on a Pulley Torque on a pulley A constant force of F = 8 N is applied to a string wrapped around the outside of a 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 Newton’s Second Law for Rotation acceleration? Simulation Why are we told that the pulley is a solid disk? 1 So we know what to use for the rotational inertia. IMR= 2 2 1 2 Torque on a pulley Two identical pulleys Solution: Simulation We take two identical pulleys, both with string wrapped around them. For the one on the left, we apply an 8 N force to ∑τv = Iαv α the string. For the one on the right, we hang an object with a 2 1 I = MR /2 weight of 8 N. Which pulley has the larger angular RF= MR2α R 2 acceleration? Why? α1 α2 1 1. The one on the left FMR= α F 2 2. The one on the right R R 3. Neither, they're equal 2F 2× () 8.0 N α == =16 rad/s2 MR (2.0 kg)× (0.5 m) F = 8N 3 mg = 8N 4 Two identical pulleys Two identical pulleys II In both cases, the driving force is 8N. In the case on the left- Suppose the pulleys have mass Mp = 2 kg, radius, R = 0.5 m. hand-side (LHS), the entire 8N would be used to accelerate We already found that the angular acceleration of the pulley the rotation of the pulley. But in the case on the RHS, part of in the case on the LHS is 16 rad/s2. Find that in the case on the 8N will be used to accelerate the block. So the pulley on the RHS. Use g = 10 m/s2. the LHS will accelerate faster. α Solution It’s a good practice to 2 2 R label the I = ½ MpR = ½ x 2kg x (0.5m) T = 0.25 kgm2 directions of α and a Applying the Newton’s eqn. to T in the the pulley, we get: τ = TR = Iα a free-body- diagrams. Applying the Newton’s eqn. to the block, we get: 8N – T = ma mg = 8N 5 6 1 Two identical pulleys II Two identical pulleys II From the above discussions, we α α It’s a good It’s a good In the above two equations, have the three equations: practice to practice to there are three unknowns, R label the R label the 2 TR = Iα = ½ M R α …….…...(1) T directions namely T, α and a. With that, T directions p 8N – T = ma …………(2) of α and a we need three equations to of α and a T in the solve for their values. The third T in the a = Rα ……………….(3) free-body- free-body- a equation comes from the fact a diagrams. that the pulley and the block are diagrams. Substitute (3) in (2), connected by the string that 8N – T = mRα ……………….(4) mg = 8N wraps around the outer rim of mg = 8N the pulley. With that, a = Rα. Eq. (1) ⇒ T = ½ MpRα …….…...(1’) (4) + (1’) ⇒ 8N = (½ Mp+m)Rα ⇒ α = 8N / [(1/2 x 2kg + 0.8kg) x 0.5m)] = 8.89 rad/s2. 7 This is less than that in case 1 as we predicted. 8 Atwood’s machine Re-analyzing the Atwood’s machine Atwood’s machine involves one pulley, and two objects When we analyzed Atwood’s connected by a string that passes over the pulley. In machine in the past, we general, the two objects have different masses. neglected the mass of the pulley (i.e., we assumed that the pulley is massless). If we include the mass of the pulley, we should expect the acceleration of the a masses m and M to be: a 1. larger than before. 2. smaller than before. 3. the same as before. 9 10 Re-analyzing the Atwood’s machine Acceleration in an Atwood’s machine II Problem: Find an expression for the acceleration of m and In an Atwood’s machine, the driving force is (M M in an Atwood’s machine with a pulley mass of m – m)g. When the pulley is not massless, part of p. the driving force is diverted into accelerating the pulley’s rotation. Solution: The pulley can be considered a solid disc, 2 with I = ½ mpR , where R is the radius of the pulley. Physical picture: In the Atwood’s machine, the tension force pulling on the heavier mass M is larger than that pulling on the lighter mass m. This results in a net clockwise torque and hence a clockwise angular acceleration in the pulley. 11 12 2 Acceleration in an Atwood’s machine II Acceleration in an Atwood’s machine II Step 1: Analyze the lighter block Step 2: Analyze the heavier block Sketch the free-body diagram for the lighter block. Sketch a free-body diagram for the heavier object. Choose a positive direction, and apply Newton’s Second Choose a positive direction, and apply Newton’s Second Law. Law. Choose positive down this time, to match the object’s Let’s choose positive to be up, i.e., in the direction of acceleration. the acceleration. FT2 FT1 F F v v v v T1 T2 ∑Fma= ∑FMa= a FT1 a FT2 +−Mg FT 2 =+ Ma +−FmgmaT1 =+ mg Mg 13 14 Acceleration in an Atwood’s machine II Acceleration in an Atwood’s machine II Step 3: Analyze the pulley Step 4: combine the equations Sketch a free-body diagram for the pulley. Choose a positive direction, and apply Newton’s Second Lighter block: +FmgmaT1 −=+ Law for rotation. Heavier block: +Mg−=+ FT 2 Ma Choose positive clockwise, to match the pulley’s angular 1 acceleration. Pulley: +−=+FFTT21 ma p ∑τv = Iαv 2 α ⎛⎞1 I = ½m R2 +−=+RF RF m R 2 α 1 p TT21⎜⎟ p Add the equations: ⎝⎠2 +−Mg mg =+++ Ma ma mp a FT1 FT2 equal 2 ⎛⎞1 a +−=+FF mR TT21⎜⎟2 pR ⎛⎞1 ⎝⎠ +−Mg mg = M ++ m m a 1 Previous result for ⎜⎟2 p +−=+FF ma ⎝⎠ TT21 p massless pulley 2 Mg− mg (Note that R cancels out so it is not Mg− mg a = a = mp needed to be specified.) 15 Mm+ Mm++ 16 2 The end 17 3.
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