Understanding Rotational Dynamics: Moment of Inertia & Torque

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University of North Texas**We aren't endorsed by this school

Course

PHYS 1710

Subject

Mechanical Engineering

Date

Dec 10, 2024

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Uploaded by PromiseOKesh

1710 Unit 3 Learning Goals 2 Below are example problems. At the end of all the problems are solutions. 1.(Level 1) A 1.0 m long, very light (essentially massless) rod has a 500 gram point mass attached 40 cm from the left end and a 700 gram point mass attached 65 cm from the left end. a) Calculate the moment of inertia for rotations about the left end of the rod. b) Calculate the moment of inertia for rotations about the center of the rod. 2.(Level 1) A 5 kg solid sphere has a moment of inertia of 1.28 kg*m2when rotating around its center of mass. The radius of the sphere is equal to: a) 50 cm b) 60 cm c) 70 cm d) 80 cm Learning Goal 2 - Rotational Dynamics2.1Calculate moment of inertia for a collection of point masses and/or a basic shape listed on the equation sheet rotating around a fixed axis.2.2Qualitatively compare moment of inertia for an object rotating around different axes or for different shapes/mass distributions.2.3Calculate the magnitude and direction of the net torque exerted on an object rotating about a fixed axis.2.4Qualitatively compare torques due to forces applied at different locations and different angles. Apply Newton’s 2nd law for rotation to qualitatively explain rotation in different scenarios.2.5Apply Newton’s 2nd law of rotation to relate torque, moment of inertia, and angular acceleration in scenarios including:-Object in static equilibrium-Rolling or rotating object being acted on by multiple forces-System of objects involving a pulley with mass-Determining the maximum applied force or maximum incline for which an object will roll without slipping

3.(Level 2) The object below consists of point masses connected by massless rods. All masses are in the xy plane. a) Calculate the moment of inertia for rotations around the z axis. b) Calculate the moment of inertia for rotations around the x axis. 4.(Level 2) Consider a pulley consisting of two concentric rings: a ring with a 1.5 kg mass and a 30 cm radius connected to a 2.2 kg ring with a 50 cm radius. The rings are connected by very light rods so that they rotate as a single object. Calculate the moment of inertia of the pulley for rotations about its center of mass. 5.(Level 1) Moment of inertia is larger when the mass is concentrated: a) close to the axis of rotation b) far from the axis of rotation

6.(Level 2) Compare the moments of inertia for the uniform ring rotating around the y axis vs rotating around the z axis. Include a one sentence justification for your answer. a) b) c) 7.(Level 2) The object below contains four identical masses connected by massless rods. The masses are arranged in a square centered on the origin. Which choice below correctly ranks the moments of inertia for rotations about the x, y, and z axes? a) b) c) d) 8.(Level 2) Suppose you are making a pizza from scratch. You start out with a uniform disc of dough and as you toss and shape the pizza you gradually end up with more dough around the edge of the pizza to form the crust. What, if anything, happens to the pizza’s moment of inertia during this process? Explain your reasoning. Iy<IzIy=IzIy>IzIx=Iy=IzIx=Iy<IzIz<Ix=IyIx<Iy<Iz

9.(Level 1) The diagram below shows a very light rod with three forces exerted on the rod. The black dot at the left end of the rod represents the axis of rotation. . a) Determine the magnitude and direction of the torque produced by . b) Determine the magnitude and direction of the torque produced by . c) Determine the magnitude and direction of the torque produced by . 10. (Level 1) The diagram below shows a very light rod being acted on by four forces. The black dot represents the axis of rotation. For each force, give the direction of the torque produced by that force. Force A: Force B: Force C: Force D: |⃗F1|=|⃗F2|=|⃗F3|= 10 N⃗F1⃗F2⃗F3

11. (Level 2) A uniform meter stick that weighs 0.8 N is subject to three applied forces as shown below. a) Calculate the magnitude of the net torque about axis A. b) Calculate the magnitude of the net torque about axis B. 12. (Level 2) A force is applied at . Calculate the magnitude and direction of the resulting torque about the z axis. 13. (Level 3) A uniform 2 kg disc with a 10 cm radius is attached to the wall through an axis that is off-center as shown below. a) When the disc is released, it will begin to rotate around the axle. Calculate the magnitude of the torque the moment the disc is released. b) As the disc begins to rotate, will the torque magnitude increase, decrease, or remain the same? Explain. 14. (Level 1) Choose the correct phrases in brackets to complete the sentence. To achieve maximum torque, you should exerts you force [close to/far from] the axis of rotation and the direction of your force should be [parallel/perpendicular]to the line from the axis of rotation to the point where the force is applied. ⃗F= (3̂i−2̂j) N⃗r= (6̂i+ 2̂j) m

15. (Level 2) We have all had the misfortune of accidentally pushing on a door near the hinges rather than near the free end and finding that we are not strong enough to open the door. Explain why the door opens when we push near the free end but not near the hinges. 16. (Challenge) An electric motor uses magnetic force to produce torque on a loop of wire carrying current. Consider an electric motor consisting of a rectangular loop of metal free to rotate around the vertical axis shown below. The magnitude of the magnetic force on the left and right sides of the loop is directly proportional to the length and independent of the width . There is no magnetic force exerted on the top and bottom sides of the rectangle. a) How will the rotational acceleration change (if at all) if we double while stays the same? b) How will the rotational acceleration change (if at all) if we double while stays the same? *Note that the loop is NOT massless 17. (Level 1) A uniform disc with a mass of 10 kg and a radius of 1.5 m is free to rotate about its center of mass. The disc experiences a net torque of 15 N*m. Calculate the magnitude of the disc’s angular acceleration. LWLWWL

18. (Level 2) A uniform disc with a mass of 1.8 kg and a radius of 2.4 m is free to rotate about its center. The disc experiences three forces as shown below. a) If the disc starts at rest, how long will it take the disc to complete two full rotations? b) Will the disc rotate clockwise or counter clockwise? c) Calculate the magnitudes of the angular, tangential, and total acceleration of point A after two full rotations. 19. (Level 2) A uniform disc with a mass of 2 kg and a radius of 40 cm has a string attached at its center of mass. The disc is initially at rest on a level floor with coefficients of friction . The string is pulled horizontally with 4 N of force. a) Calculate the angular acceleration of the disc. b) Calculate the friction force exerted by the floor. μk= 0.1 andμs= 0.25

20. (Level 3) A 1.5 m long non-uniform sign with a mass of 80 kg is held in static equilibrium by three cables as shown below. Determine the tension in cables 1 and 3 and the location of the sign’s center of mass. 21. (Level 3) Two boxes are connected by a massless rope passing over a 4 kg hollow ring acting as a pulley that can rotate about its center without friction. Determine the tension in each side of the rope and the magnitude of angular acceleration of the pulley while the 8 kg box falls.

22. (Level 3) A 3 kg solid disc is released from rest at the top of an inclined plane with coefficients of friction . a) Calculate the linear acceleration of the disc as it rolls down the incline if . b) Calculate the largest value of for which the disc will roll without slipping (i.e. at any larger angle the disc will star too slip as it moves down the incline). μs= 0.2 andμk= 0.1θ= 15oθ

Answers (solutions below) 1.0.376 kg*m2and 0.021 kg*m2 2.D 3.39.2 kg*m2and 9.2 kg*m24.0.685 kg*m25.B 6.A 7.B 8.See solution 9.4 N*m out of the page, 3.44 N*m into the page, 0 N*m 10.Into the page, out of the page, out of the page, into the page 11.1.4 N*m and 3.5 N*m 12.18 N*m 13.1.18 N*m, decrease 14.Far from, perpendicular 15.See solution 16.Double, cut in half 17.1.33 rad/s218.11.66 s, CCW, 10.26 m/s2, 0.407 m/s2, 10.27 m/s219.3.33 rad/s2, 1.33 N 20.774.56 N, 28.38 N, 0.028 m to the right of rope 1 21.57.65 N, 64.57 N, 1.73 m/s222.2.54 N, 1.69 m/s2, 31o

Solutions 1.A 1.0 m long, very light (essentially massless) rod has a 500 gram point mass attached 40 cm from the left end and a 700 gram point mass attached 65 cm from the left end. a) Calculate the moment of inertia for rotations about the left end of the rod. b) Calculate the moment of inertia for rotations about the center of the rod. 2.(Level 1) A 5 kg solid sphere has a moment of inertia of 1.28 kg*m2when rotating around its center of mass. The radius of the sphere is equal to: a) 50 cm b) 60 cm c) 70 cm d) 80 cm3.The object below consists of point masses connected by massless rods. a) Calculate the moment of inertia for rotations around the z axis. For point masses we can use where r is measured from the rotation axis. For the z axis (into and out of the page) these distances are simply the lengths of the massless rods. I=∑mir2i= (0.5 kg)(0.4 m)2+ (0.7 kg)(0.65 m)2= 0.376 kg⋅m2I=∑mir2i= (0.5 kg)(0.1 m)2+ (0.7 kg)(0.15 m)2= 0.021 kg⋅m2Isphere,cm=25MR2→1.28 kg⋅m2=25(5 kg)R2I=∑mir2i

b) Calculate the moment of inertia for rotations around the x axis. The 5 kg is on the x axis and the 3 kg and 7 kg are closer to the x axis than they are the z axis so we expect a smaller moment of inertia. Extension:Suppose each rod was a uniform rod with a 1 kg mass. How would you factor that into your calculations? 4.Consider a pulley consisting of two concentric rings: a ring with a 1.5 kg mass and a 30 cm radius connected to a 2.2 kg ring with a 50 cm radius. The rings are connected by very light rods so that they rotate as a single object. Calculate the moment of inertia of the pulley for rotations about its center of mass. 5.Moment of inertia is larger when the mass is concentrated: a) close to the axis of rotation b) far from the axis of rotationThis is what the equation says. For a given collection of masses the larger the the larger the moment of inertia value. The is the distance of each mass from the axis of rotation. We can also see this result when we compare the moment of inertia formula for a solid sphere vs a spherical shell. For the same total mass the shell has a larger moment of inertia because its mass is concentrated around the edge far from the axis of rotation. Iz= (5 kg)(1.5 m)2+ (3 kg)(1.6 m)2+ (7 kg)(1.7 m)2= 39.2 kg⋅m2Ix= (5 kg)(0 m)2+ (3 kg)[(1.6 m)sin(56o)]2+ (7 kg)[(1.7 m)cos(64o)]2= 9.2 kg⋅m2Ipulley=Iring1+Iring2= (1.5 kg)(0.3 m)2+ (2.2 kg)(0.5 m)2= 0.685 kg⋅m2I=∑mir2iriri

6.Compare the moments of inertia for the uniform ring rotating around the y axis vs rotating around the z axis. Include a one sentence justification for your answer. a) b) c) For rotations around the y axis the mass near the top and bottom of the ring is close to the axis of rotation. For rotations around the z axis (perpendicular to the page) all of the ring’s mass is far from the axis of rotation. The more mass that is far from the axis of rotation, the larger the moment of inertia. 7.The object below contains four identical masses connected by massless rods. The masses are arranged in a square centered on the origin. Which choice below correctly ranks the moments of inertia for rotations about the x, y, and z axes? a) b) c) d) Each mass is a distance a/2from the x axis and a distance a/2from the y axis so . Each mass is a distance (Pythagorean theorem) from the z axis which is further than a/2and the further mass is from the rotation axis, the larger the moment of inertia. Iy<IzIy=IzIy>IzIx=Iy=IzIx=Iy<IzIz<Ix=IyIx<Iy<IzIx=Iya/2

8.Suppose you are making a pizza from scratch. You start out with a uniform disc of dough and as you toss and shape the pizza you gradually end up with more dough around the edge of the pizza to form the crust. What, if anything, happens to the pizza’s moment of inertia during this process? Explain your reasoning. Mass is moving towards the edge of the disc which is further from the rotation axis (the center of the disc) so moment of inertia is increasing. Note that the size of the disc is not changing, but mass is moving towards the outside of the disc which we can reason by or by comparing for a solid sphere and a spherical shell will increase moment of inertia. 9.The diagram below shows a very light rod with three forces exerted on the rod. The black dot at the left end of the rod represents the axis of rotation. . a) Determine the magnitude and direction of the torque produced by . , out of the page b) Determine the magnitude and direction of the torque produced by . , into the page c) Determine the magnitude and direction of the torque produced by . I=∑mir2iI|⃗F1|=|⃗F2|=|⃗F3|= 10 N⃗F1|⃗τ|=|⃗r||⃗F|sin(θ) = (0.4 m)(10 N)sin(90o) = 4 N⋅m⃗F2|⃗τ|=|⃗r||⃗F|sin(θ) = (0.6 m)(10 N)sin(35o) = 3.44 N⋅m⃗F3|⃗τ|=|⃗r||⃗F|sin(θ) = (0.8 m)(10 N)sin(0o) = 0 N⋅m

10. The diagram below shows a very light rod being acted on by four forces. The black dot represents the axis of rotation. For each force, give the direction of the torque produced by that force. Force A: into the page Force B: out of the page Force C: out of the page Force D: into the page Right hand rule: pointer finger from axis of rotation to where force is exerted (the point along the dashed line where each arrow is drawn), point middle finger in the direction the arrow points (without moving your pointer finger), thumb points in the direction of torque. Only the perpendicular component of force produces torque (you can see this in the calculations from the previous problem). This means that for the right hand rule if your force points at an angle, you can point your middle finger in the direction of the perpendicular component of force (perpendicular to the vector from axis of rotation to where force is exerted). 11. A uniform meter stick that weighs 0.8 N is subject to three applied forces as shown below. ⃗r0.8 N

a) Calculate the magnitude of the net torque about axis A. When adding torques, I like to skip the right hand rule and instead just think about the direction each individual force tries to rotate the object to determine whether the torques should add or subtract. Imagine pinching the stick at the axis and applying each force individually and think about (or try with your pencil) what direction the stick rotates. For axis A: 2 N tries to rotate the stick clockwise, 1N does nothing (r = 0), 4 N tries to rotate counterclockwise. The weight of the stick is exerted at the center of mass and the center of mass is at 50 cm because the stick is uniform. This means the 0.8 N weight produces no torque. b) Calculate the magnitude of the net torque about axis B. For axis B: 2 N tries to rotate counterclockwise, 1 N tries to rotate clockwise, 4 N tries to rotate counterclockwise. Now the 0.8 N weight at 50 cm does produce torque and that torque tries to rotate the stick clockwise. Note:Make sure you understand why the 2 N and 4 N forces produce opposite direction torques in a) but the same direction torques in b). This can be very confusing when first studying torque. Note:If you want to know what direction the rod will rotate, just remove the absolute value signs and look at whether the net torque is positive or negative. We wrote counterclockwise torques as positive so a positive net torque would mean counterclockwise rotation. Extension: Suppose you apply a 3rd force at the 20 cm point on the stick. Calculate the magnitude and direction this force must have to make the net torque = 0. The fact that you get the same answer for axis A and axis B shows that you can choose any point as your axis of rotation when solving static equilibrium problems. 12. A force is applied at . Calculate the magnitude and direction of the resulting torque about the z axis. We can use the cross product to calculate torque or we can use dot product to find the angle between and use with the right hand rule to find direction. I will show the first option, you can try the second option on your own. τ=|−(0.1 m)(2 N) + (0 m)(1 N) + (0.4 m)(4 N)|= 1.4 N⋅mτ=|(0.4 m)(2 N)−(0.5 m)(1 N) + (0.9 m)(4 N)−(0.5 m)(0.8 N)|= 3.5 N⋅m⃗F= (3̂i−2̂j) N⃗r= (6̂i+ 2̂j) m⃗rand⃗F|⃗τ|=rFsinθ

So the torque about the z axis (the z component of our answer) has a magnitude of 18 N*m and the direction is in the negative z direction. 13. A uniform 2 kg disc with a 10 cm radius is attached to the wall through an axis that is off-center as shown below. a) When the disc is released, it will begin to rotate around the axle. Calculate the magnitude of the torque the moment the disc is released. To do this calculation we need to know that for any object we can take the gravitational force to act at the center of mass and we need to know that referring to an object as “uniform” (uniform density) means that the center of mass is the center of the object. b) As the disc begins to rotate, will the torque magnitude increase, decrease, or remain the same? Explain. The magnitude will decrease because the angle between and the gravitational force is no longer 90 degrees once the disc starts to rotate. ⃗τ=⃗r×⃗F=̂îĵk6203−20= 0̂i−0̂j+ (−12−6)̂kτ=rFsinθ= (0.06 m)(2 kg)(9.8 N/kg)sin(90o) = 1.18 N⋅m⃗r

14. Choose the correct phrases in brackets to complete the sentence. To achieve maximum torque, you should exerts you force [close to/far from] the axis of rotation and the direction of your force should be [parallel/perpendicular]to the line from the axis of rotation to the point where the force is applied. exerting force far from the rotation axis makes large. Exerting the force perpendicular to the line connecting the axis to where the force is applied makes . 15. We have all had the misfortune of accidentally pushing on a door near the hinges rather than near the free end and finding that we are not strong enough to open the door. Explain why the door opens when we push near the free end but not near the hinges. In order to open the door we need to produce a noticeable angular acceleration. The moment of inertia of the door is the same no matter where the push but the torque is not. The rotation axis is the edge with the hinges so pushing near the hinges produces a small torque (and small ) while pushing near the free end produces a large torque (and large ). 16. An electric motor uses magnetic force to produce torque on a loop of wire carrying current. Consider an electric motor consisting of a rectangular loop of metal free to rotate around the vertical axis shown below. The magnitude of the magnetic force on the left and right sides of the loop is directly proportional to the length and independent of the width . There is no magnetic force exerted on the top and bottom sides of the rectangle. |⃗τ|=|⃗r||⃗F|sinθ|⃗r|θ= 90oα=τIα=τILW

a) How will the rotational acceleration change (if at all) if we double while stays the same? Angular acceleration doubles. If we double the force and thus the torque will double while the moment of inertia stays the same (the distance of mass from the rotation axis has not changed). This will double the angular acceleration. b) How will the rotational acceleration change (if at all) if we double while stays the same? Angular acceleration is cut in half. If we double the force stays the same but the distance from the rotation axis to the force doubles so torque will double. However, the moment of inertia of the loop will quadruple (distance from axis to the mass is squared in ) so the angular acceleration will be cut in half. 17. A uniform disc with a mass of 10 kg and a radius of 1.5 m is free to rotate about its center of mass. The disc experiences a net torque of 15 N*m. Calculate the magnitude of the disc’s angular acceleration. 18. A uniform disc with a mass of 1.8 kg and a radius of 2.4 m is free to rotate about its center. The disc experiences three forces as shown below. LWLWLWIIdisc,cm=12MR2=12(10 kg)(1.5 m)2= 11.25 kg⋅m2τnet=Iα→α=15 N⋅m11.25 kg⋅m2= 1.33 rad/s2

a) If the disc starts at rest, how long will it take the disc to complete two full rotations? b) Will the disc rotate clockwise or counter clockwise? Counterclockwise. CCW is what I used as positive for torque directions and the net torque cam out positive. c) Calculate the magnitudes of the centripetal, tangential, and total acceleration of point A after two full rotations. Centripetal acceleration Tangential acceleration Total acceleration 19. A uniform disc with a mass of 2 kg and a radius of 40 cm has a string attached at its center of mass. The disc is initially at rest on a level floor with coefficients of friction . The string is pulled horizontally with 4 N of force. a) Calculate the angular acceleration of the disc. b) Calculate the friction force exerted by the floor The 4 N force is trying to slide the disc to the right so the disc will experience a Idisc,cm=12MR2=12(1.8 kg)(2.4 m)2= 5.18 kg⋅m2∑τ= (2.0 m)(15 N)sin(35o)−(1.0 m)(6 N)sin(45o)−(1.2 m)(10 N)sin(90o) = 0.96 N⋅m∑τ=Iα→(0.96 N⋅m) = (5.18 kg⋅m2)α→α= 0.185 rad/s2Δθ=ωiΔt+12αΔt2→4π= 0 +12(0.185)Δt2→ Δt= 11.66 sω= (0.185 rad/s2)(11.66 s) = 2.16 rad/sv=ωR= (2.16 rad/s)(2.2 m) = 4.75 m/sac=v2R=(4.75 m/s)22.2 m= 10.26 m/s2at=αR= 0.407 m/s2at=a2c+a2t= 10.27 m/s2μk= 0.1 andμs= 0.25fs

friction force to the left at the point of contact between the disc and the ground. If the disc rolls without slipping, this will be static friction (but we cannot assume it will be the maximum static friction!). If the disc rolls without slipping, that means . Making this substitution and solving the system of equations gives and . Before calling it a day, we have to check that this friction force is possible. . The calculated friction value is less than the maximum value so our answer is correct. Note:Usually I write torques that try to cause clockwise rotation as negative. In this problem I wrote the frictional torque as positive because rotating clockwise moves the disc to the right and “to the right” is what I called the positive direction for force. You want your positive directions for torque and force to agree. This idea becomes essential if the 4 N force was exerted somewhere else and you needed to determine the sign of each torque. Extension:Determine the largest pulling force (> 4 N) that can be exerted before the disc starts to slip. Suppose the 4 N force was exerted 20 cm above or below the center of the disc. Calculate the new friction and angular acceleration for that scenario. 20. A 1.5 m long non-uniform sign with a mass of 80 kg is held in static equilibrium by three cables as shown below. Determine the tension in cables 1 and 3 and the location of the sign’s center of mass. ∑F=ma→(4 N)−fs= (2 kg)a∑τ=Iα→(0.4 m)fssin(90o) =12(2 kg)(0.4 m)2αa=αRfs= 1.33 Nα= 3.33 rad/s2fs,max=μsFN= 0.25(2 kg)(9.8 N/kg) = 4.9 N

Three ideas we need for this problem 1) static equilibrium means sum of forces in each direction = 0 and sum of torques = 0, 2) we can choose any location we want as the axis of rotation, and 3) even though we don’t know where the center of mass is, we can still mark it in the diagram and label the distance from our axis of rotation to the center of mass as x to allow us to write equations. We can solve the first equation for T3then plug that into the second equation to solve for T1 then plug both of those into the third equation to solve for x. center of mass is 0.028 m from the axis of rotation labeled above. 21. Two boxes are connected by a massless rope passing over a 4 kg hollow ring acting as a pulley that can rotate without friction. Determine the tension in each side of the rope and the magnitude of angular acceleration of the pulley while the 8 kg box falls. Starting ideas: 1) the tension in each side of the rope will be the forces that produce torque on the pulley and 2) we can use to relate the angular acceleration of the pulley to the linear acceleration of the boxes. I will use an x axis that curves around the pulley. I need my angular acceleration directions to agree with my linear acceleration directions so this means that torque that tries to rotate the pulley clockwise is positive and torque that tries to rotate the pulley counterclockwise is negative. ∑Fx= 0→ −22 cos(34o) +T3cos(50o) = 0∑Fy= 0→T1−22 sin(34o)−(80)(9.8) +T3sin(50o) = 0∑τ= 0→(0)T1+ (0.15)22 sin(34o) +x(80)(9.8)sin(90o)−(1.1)T3sin(50o) = 0T1= 774.56 N,T3= 28.38 Nα=a/r∑5kgF=Ma→TL−(5)(9.8) = 5a∑8kgF=Ma→(8)(9.8)−TR= 8a∑τ=Iα→(0.6)(TR)−(0.6)(TL) = (4)(0.6)2a0.6

Solving this system of equations we get , , and . Note:You cannot treat the two boxes as one system the way we sometimes do in Unit 2 because the pulley with mass causes the tension to be different on each side of the rope. The system equation we used in Unit 2 is what we would get from adding together the two force equations IF the two tensions were equal. Note:Many students don’t think about relating their linear acceleration and angular acceleration directions. If those directions are not consistent, you will get an incorrect answer. Read through the 2nd comment above the math again and think about how we match those directions. Note: This problem is easier to solve using conservation of energy (let the boxes move a given distance then use CoE to calculate and from that calculate then look at each box individually to find tension values). I encourage you to try that solution. Extension:Notice that the radius of the ring cancels out. Some students get stuck if they are not told the radius not working far enough to see it cancels out. How would the solution change if the 5 kg box was supported on a flat surface with friction? How would the tension or acceleration values change if we use a solid disc as a pulley instead of a hollow ring - try to answer conceptually without redoing all the calculations. 22. A 3 kg solid disc is released from rest at the top of an inclined plane with coefficients of friction . a) Calculate the linear acceleration of the disc as it rolls down the incline if . TL= 57.65 NTR= 64.57 Na= 1.73 m/s2vaμs= 0.2 andμk= 0.1θ= 15o

We need to write and for the disc and set as the condition for rolling without slipping. The rotation axis is the center of the disc. Friction force and the normal force are applied at the radius of the disc where the disc contacts the ramp while gravitational force is applied at the center of the disc. (The question should say that the disc is uniform.) Solving these together gives . b) Calculate the largest value of for which the disc will roll without slipping (i.e. at any larger angle the disc will star too slip as it moves down the incline). To find the maximum angle we have the same equations as above except we replace with the max static friction value. Solving these together (I did this by solving each equation for then graphing those two expressions to see where they intersect) gives . Extension: The disc is moving - how do you explain why we use static friction not kinetic friction? If the disc starts to slip, will be larger or smaller than ? (use a prop and act it out to help you visualize) ∑F=ma∑τ=Iαα=aR∑Fx=max→Mgsin(15o)−fs=Max∑τ=Iα→R fs=12MR2axRfs= 2.54 N anda= 1.69 m/s2θfs∑Fx=max→Mgsinθmax−μsMgcosθmax=Max∑τ=Iα→RμsMgcosθmax=12MR2axRaxθmax= 31oanda= 3.36 m/s2αaxR