Question

Make sure to consider all parts of the problem with detail and accuracy. Please show your work and explain. Thank you!

A cylindrical wheel of uniform density with mass M=11kg and radius R= 2m rolls without slipping on an incline that makes an angle of 0 = 30° with the horizontal, as shown in the diagram; there is friction between the wheel and the ramp. A block of equal mass Mis sliding on a separate, frictionless incline, also forming an angle of 0 = 30° with the horizontal. A thin, ideal rope tied to the block passes over an ideal pulley near the peak between the two ramps, and wraps around the outside of the rolling wheel; the rope is parallel to the ramp on each side of the apparatus. Recall that the moment of inertia of a cylinder about its axis of symmetry is Icm = 42 MR2. Recall that sin(30°) = 12 and cos(30°) ~ 0.9. a) Draw 2 separate “extended” free body diagrams, one for the block and one for the wheel, indicating where every force acts on each of these objects. b) What is the magnitude of the normal force acting on the wheel? c) If the wheel were to roll a distance of 1 m down the ramp, then how far would the block slide during that time? M d) What is the acceleration of the block? Clearly friction indicate whether it accelerates up or down the incline. M no R7 friction e e) What is the tension in the rope?

Answer 1

solution F. B.D of block Ni Let block is sliding up with accoleration a'. 11 30' 30 using Necuton's and law of 11 g Sinzo' 119 motion 30° N = 119 Cossol 11g sin 30 = lla F. Bidig wheel N₂ using Newton's and law of motion ng sino' lng coso 18 30° N = 11g in Los 30' lig sino' (T + dk ) = lla 44 resultant of As the force of friction J & tension in rope provides the necessary torque for rolling. Therefore (T-f R = I a = = I MRV) (T-2) = I mi? Com T = J = Ima ET - Ima = T - la 7 = T - 5.5a 2 {:'M=11kg 2 5 Y C Scanned with CamScanner

Now from eqh 11 g Sin 30' (T & T - 5,50) = lla 6 11 g Sinzo - 2T = 11a - 5.5a 11g sin 30 - 2T = 5.5a by 2' multiplying eqr adding with eah (2T - 22g sin 30') + (11g sin30 - 2T) = 22a +5.5a - 11g sin 30' = 27.5a = 27.59 - 1 x 9.8 x 1 10718 55 er a= -119.8 2x27.5 Part 2 Here and for the acceleration (1.96 m/s2 a = - 1.96 m/s2 ve shows wheel roll up the in ineline block accelerates down with Ang Now from eah @ T - 119 Sinso = 1 (-1196) Ta 1189.8 x 1 - 21:56 T - 53.7 - -21.56 T = -21.56+ 53.9 which ở tension T = 32 34 N ~ 32 N Bay from eah N2 119 Cos 30 = 11X 9.88 0.9 2 іn уоре . ginen con 30-0.9 = 97.02 N ~ 97 N which is normal force acting on wheel. Pres. CS Scanned with CamScanner

For wheel, using equation of motion a = bit - I at2 a = 1.96 m/s2 Here, af = m I Fot7 X 1.9642 t² = 2XL 2 = 1.020 1.96 1.96 taking square root on both side tools For block s = Vit taat? f = Otx1.96 * (lony s = 0.999 m ~ im block will also slide 1 m during Buy 1 Thus that time: CS Scanned with CamScanner
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