Applications - Frictions and Power Transmission (October 11, 2024) PDF

10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review Started on Monday, 21 October 2024, 1:33 AM State Finished Completed on Monday, 21 October 2024, 2:17 AM Time t...

10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review Started on Monday, 21 October 2024, 1:33 AM State Finished Completed on Monday, 21 October 2024, 2:17 AM Time taken 44 mins 20 secs Grade 1.0 out of 30.0 (3.3%) Feedback This activity was based on previous teamwork! Please take time to review your solution in the light of the following video resources. Please click, watch , wiser, … Video 1 Video 2 https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 1/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review Question 1 Partially correct Mark 1.0 out of 15.0 A rectangular paddle (L × h) is attached to the driven pulley of the open belt-drive system illustrated in the figure. The belt-drive subsystem is driven by friction plate. The rectangular paddle has a neglibile moment of inertia Ip and radius of inertial kp. A normal force of Fn = 600 N presses the driven plate against driven plate and the coefficient of friction between the two is μ1 = 0.32. Assumming that only one side of the friction plate serves in the power transmission under equal pressure distribution , you are required to analyse the system and evaluate the net power at the driven pulley and the driver pulley as well as the acceleration of the system from rest. The driver plate has an internal diameter of ϕi = 26 cm and external diameter ϕo = 40 cm. The driven plate of diameter ϕo = 40 cm is attached to the driver pulley and both are solid cylindrical parts of mass m1 = 10 kg each and the driver pulley has a radius of 130 mm. The driven pulley has an external diameter of 47 cm with an inner diameter of 40 cm with a mass of 9 kg. The mass of spokes connecting the pulley to the hub is negliglible. The coefficient of friction between the belt and both pulleys is 0.3. The axes of both pulleys are 0.54m away and the belt has a total mass of 1.8 kg and the tip of the mixing paddle is 26.5 cm away from the centre of rotation. Analyze the power transmission system with detailed diagrams of moments and forces, equations in order to evaluate: 1. The acceleration of the system (from rest), 2. Torques and tensions, and 3. Conditions of optimal power transmission for the system It is establised that the following are required for the solution of the problem The pressure intensity at a points located at at r = 20 cm from the axis of rotation of friction plate : p(20 cm) = 8267.789 N/m2  One possible correct answer is: 8267.789251527 The input torque that powers the pulleys side of the transmission system has a magnitude of 38.4 N/m2  One possible correct answer is: 32.155151515152 The moment of Inertia of the combined (driven plate and driver pulley) is I1 = 0.456 kgm2 (*)  One possible correct answer is: 0.2845 A bit of theory check: decide which one of these options is correct for the computation of the moment of inertia of the driven pulley of mass mp with an outer radius Ro and inner radius Ri given, R2o + R2i R2o − R2i I = mp (a) or I = mp (b) ; following proper theory-based analysis, the correct option is option b 2 2  One possible correct answer is: option a The radius of inertia of the driven pulley is 166.67 mm https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 2/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review  One possible correct answer is: 218.20288724029 After proper analysis of the geometry of the belt arrangement, one can establish that the belt wraps both pulleys over angles of θ1 = 1.466 rad θ2 = 1.466 o   One possible correct answer is: 2.7502105478593 at the smaller one and One possible correct answer is: 202.4245428353 for the bigger one and then calculate: The linear acceleration of the driver pulley: m/s2  One possible correct answer is: 7.2583942311738 The angular acceleration of the tip of the mixing paddle: rad/s2  One possible correct answer is: 30.886783962442 The tension in the tight side of the belt is N  One possible correct answer is: 154.90243151102 If the maximun power transmitted with a belt of λ kg/m of mass per unit legnth of the belt is acheived at a speed a speed v such that 1 λv2 = Fmax. Assuming that the tension in the tight side of the belt calculated above is the maximum the belt can take, calculate the 3 maximum speed vmax = m/s. The total mass of the belt is 1.8 kg (*)  One possible correct answer is: 13.915029262951 The optimal speed is attained sec after the clutch engages the driven plate (*)  One possible correct answer is: 1.9170947209216 Your answer is partially correct. You have correctly answered 1 part(s) of this question. The story: Clutch plate from motor: To − Tμ1 = Io αo 2 r3o − r3i ro + ri with Tμ1 = μ1 Fn 2 or T mu1 = μ1 Fn 3 ro − r2i 2 Speculate about the pressure distribution : pmin and pmax Driven plate & driver pulley Tμ1 − F1 ⋅ r1 + F2 ⋅ r1 = I1 α1 F1 − Fc with = eμ2 θ and Fc = λv2b (Variable speed: make sure the optional speed is not exceeded !!! - ignore the centrifugal F2 − Fc component and only incorporate it ate a later stage of the problem.) https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 3/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review Calculation of angle of contact : θ1 = and θ2 Driver Pullye / Belt: F1 ⋅ r2 − F2 ⋅ r2 = I2 α2 End of story the rest is mathematical logical sense r1 α1 = r2 α2 assuming no slippage between the belt and pulleys. F2 = F1 ⋅ e−μ2 θ Tμ1 − F1 ⋅ r1 (1 − e−μ2 θ ) = I1 α1 and F1 ⋅ r2 (1 − e−μ2 θ ) = I2 α2 Equations to be solved for F1 and accelerations α1 and α2. Possible approach solving both equations for Tμ1 − I1 α1 I2 α2 F1 then determining the accelerations : − = F1 = r1 ⋅ (1 − e μ2 θ ) r2 ⋅ (1 − e−μ2 θ ) Tμ1 ⋅ r22 Resulting into: α1 = I1 ⋅ r22 + I2 ⋅ r21 Δω1 I1 = Idisc + Icyl and I2 = Idisc2 + 3 ⋅ Irec. The rate of change of the speed. α1 = Δt Numerical results The coefficient of friction between the driver plate and the driven plate is μ1 = 0.32 and a normal forcea Fn = 600 N presses both plates during the transmission of power to the driven plate and the belt. Assumming that only one side of the friction plate serves , calculate the magnitude of the torque transmitted trhough friction between plates: T1 = 32.155151515152 Nm or T1 = 31.68 Nm 2 0.23 − 0.133 0.2 + 0.13 T1 = ⋅ 0.32 ⋅ 600 2 Nm or T1 = 0.32 ⋅ 600 Nm 3 0.2 − 0.13 2 2 Calculate the intensity of the pressure distribution at a point For old plate k = 1364.185226502 N/m wear therefore : at r = 13 cm : pi = 10493.732511554 N/m2 at r = 16.5 cm : pm = 8267.789251527 N/m2 and at r = 20 cm : po = 6820.9261325098 N/m2 at r = 20 cm : p = 6820.9261325098N/m2 Assume that the driven plate and the driver pulley have the same mass m1 = 10 kg and that the driver pulley has a radius of 130 mm and calculate the comined moment of Inertia (driven plate and driver pulley) 2m1 ⋅ (R2o + R2i ) −−I−− = 0.2845kgm2 and the radius of gyration k1 = √ 1 I1 = = 119.26860441877 mm 2 2m1 The driven pulley has an internal diameter of 40 cm and an external diameter of 47 cm and a mass of 9 kg −−− mp ⋅ (0.272 + 0.22 ) = 0.95225kgm and the radius of gyration k2 = √ 2 I2 I2 = = 218.20288724029 mm 2 mp Tμ1 ⋅ R2p α1 = = 55.83380177826 rad/s2 ; I1 ⋅ R2p + I2 ⋅ R2i θ1 Rp − Ri 0.235 − 0.13 Angle of wrap cos = = and θ = 157.5754571647o 2 d 0.54 Tμ1 − I1 ⋅ α1 I2 αp F1 = −θ⋅μp = = 154.90243151102 N and F2 = F1 ⋅ e−θ⋅μp = 67.87937821329 N Ri ⋅ (1 − e ) Rp ⋅ (1 − e−θ⋅μp ) The net torque at the driven pulley is Rp ⋅ (F1 − F2 ) = Nm And at the driver pulley Tμ1 − Ri ⋅ (F2 − F1 ) = Nm F1 Optimal power transmission for λv2max = or vmax = 13.915029262951 m/s 3 ϕi The time to reach this speed from rest at an acceleration of α ⋅ = 7.2583942311738 m/s2 is 2 13.915029262951 t= = 1.9170947209216 s 7.2583942311738 https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 4/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review Question 2 Incorrect Mark 0.0 out of 15.0 Note: The figure is for illustration purpose only. Some values may be small given the units used. Please try and keep your work up to 5 decimal places, example 3.14159 The motor of the experimental vehicle, illustrated in the figure is driven by the oscillations of the compound pendulum. Assuming that the speed of the motor equals the angular frequency of the pendulum ω analyse the compound pendulum made of a triangular plate of height 34 cm and base 2a = 10 cm, and a semi-circular disc of radius 5 cm. The plate has a density of 157kg/m2. Point O is the tip of the triangular shape, located 34 cm above the diameter of the semi-circle. Analyse the compound pendulum and evaluate: The moment of inertia about a vertical axis passing through the centre of gravity of the triangular part: I1yyg = 0.0358 kgm2  One possible correct answer is: 1.1120833333333E-7 The moment of inertia of the triangular part about O : I1O = 0.1215 kgm2.  One possible correct answer is: 1.5538028333333E-5 The moment of inertia about a vertical axis passing through the centre of gravity of the semi-circular part: I2yyg = 0.0008 kgm2.  One possible correct answer is: 3.8533597391687E-8 The moment of inertia of the semi-circular part about O : I2O = 0.081 kgm2  One possible correct answer is: 8.0939080350165E-6 The distance between the CG (centre of gravity) of the pendulum and point O is ȳ = 0.269 m (*1)  One possible correct answer is: 0.25191598689683 The moment of inertia of the pendulum about pivot O is IO = 0.2025 kgm2  One possible correct answer is: 2.363193636835E-5 https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 5/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review 1.623  The period of oscillations of the pendulum Tp = One possible correct answer is: 1.0719238557085 s and the radius of inertia k = 0.322  One possible correct answer is: 0.26819238116954 m Evaluate the length of the equivalent simple pendulum oscillating at the same frequency Leq = 0.65 m  One possible correct answer is: 0.28552039988967 The horizontal projection of the motion of any point on the pendulum can be described by x(t) = Xm ⋅ cos(2π 2.385 ⋅ t + ϕo )  One possible correct answer is: 0.93290208504502 Calculate the speed of the motor Nmot = 22.778 rev/min  One possible correct answer is: 351.69580042753 A gear box multiplies the speed of the motor by a factor of 4 ; if the wheels have a radius of 0.35 m, the vehicle will move at linear speed 1.268 m/s 3.02 m/s2   of One possible correct answer is: 8.2062353433089 with an acceleration of One possible correct answer is: 0. The distance travelled by 2 m  the car when the pendulum swings between two extreme positions is One possible correct answer is: 4.3982297150257 and the speed of 0 m/s  the car when the pendulum is at the lowest point is One possible correct answer is: 8.2062353433089. Your answer is incorrect. You have correctly answered 0 part(s) of this question. Check : dens 7850 thick 2cm 15700 Note: the figure is for illustration purpose only. Some values may be small given the units used. Please try and keep your work up to 5 decimal places, example 3.14159 The motor of the experimental vehicle, illustrated in the figure Is driven by the motion oscillation of the compound pendulum. Assuming that the motor has the angular frequency of the pendulum ω analyse the compound pendulum made of a triangular shape of height 34 cm and base 2a = 10 cm, and a semi-circular part of part of radius 5. Analyse the compound pendulum and evaluate: {#1} The moment of inertia about a vertical axis passing through the centre of gravity of the triangular part: I1yyg = 1.1120833333333E − 7 kgm2. m1 h2 m1 a2 I1xxg = = 1.7140911111111E − 6 kgm2 or I1yyg = = 1.1120833333333E − 7 kgm2 resulting into 18 6 I1g = 1.8252994444444E − 6 kgm2 (*1) https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 6/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review 2h I1O = I1g + M1 ( )2 = 1.5538028333333E − 5 kgm2 the moment of inertia of the triangular part about O (*1) 3 And {#2} The moment of inertia about a vertical axis passing through the centre of gravity of the semi-circular part: I2yyg = 3.8533597391687E − 8 kgm2. a2 4a I2xxg = m2 − m2 ( )2 = 1.0769901763434E − 8 kgm2 or 4 3π a2 I2yyg = 2 m = 3.8533597391687E − 8 kgm2 = 3.8533597391687E − 8 kgm2 resulting into 4 I2g = 4.9303499155122E − 8 kgm2 (*1) 4a 2 I2O = I2g + M2 (h + ) = 8.0939080350165E − 6 kgm2 the moment of inertia of the semi-circular part about O (*1) 3π 2h 4a The distance between the CG of the figure and the point O is A1 + A2 (h + ) 3 3π 2h 4a A1 + A2 (h + ) ȳ = 3 3π = 0.25191598689683 m (*1) A1 + A2 The moment of inertia of the pendulum about pivot O is IO = 2.363193636835E − 5 kgm2 −−−−−− The period of oscillations of the pendulum Tp = 2π√. IO = 1.0719238557085 s (*1) and the radius of inertia Mghg −− − k = √ O = 0.26819238116954 m I M Tp 2 Evaluate the length of the equivalent simple pendulum oscillating at the same frequency Leq = g( ) = 0.28552039988967 m 2π The horizontal projection of the motion of the any point on the pendulum can be described by x(t) = Xm ⋅ cos(2π 0.93290208504502 ⋅ t + ϕo ) The horizontal motion of the x(t) = Xm ⋅ cos(… t + ϕo ) = Xm ⋅ cos(5.8615966737921 ⋅ t + ϕo ) or 2π x(t) = Xm ⋅ cos(2π … t + ϕo ) = Xm ⋅ cos(2π ⋅ (0.93290208504502) ⋅ t + ϕo ) or x(t) = Xm ⋅ cos( … t + ϕo ) or Tp --- ω Calculate the speed of the motor Nmot = 60rev/min = 351.69580042753 rev/min 2π A gear box multiplies the speed of the motor by a factor n, calculate the linear speed of the vehicle for a wheel of radius Rw : v = ω ⋅ n ⋅ Rw and the acceleration of the of the car a = 0 m/s2 --- Tp Calculate the distance travelled by the vehicle when the pendulum swings between two extreme postions ti = s 2 Tp The distance is therefore x = v and the speed when of the car when the pendulum is at its lowest point v = constant 2 Numerical Results The moment of inertia about a vertical axis passing through the centre of gravity of the triangular part: I1yyg = 1.1120833333333E − 7 kgm2. m1 h2 m a2 I1xxg = = 1.7140911111111E − 6 kgm2 or I1yyg = 1 = 1.1120833333333E − 7 kgm2 resulting into 18 6 I1g = 1.8252994444444E − 6 kgm2 (*1) 2h 2 I1O = I1g + M1 ( ) = 1.5538028333333E − 5 kgm2 the moment of inertia of the triangular part about O (*1) 3 And The moment of inertia about a vertical axis passing through the centre of gravity of the semi-circular part: I2yyg = 3.8533597391687E − 8 kgm2. a2 4a I2xxg = m2 − m2 ( )2 = 1.0769901763434E − 8 kgm2 or 4 3π a2 I2yyg = m2 = 3.8533597391687E − 8 kgm2 = 3.8533597391687E − 8 kgm2 resulting into 4 I2g = 4.9303499155122E − 8 kgm2 (*1) 4a 2 I2O = I2g + M2 (h + ) = 8.0939080350165E − 6 kgm2 the moment of inertia of the semi-circular part about O (*1) 3π https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 7/8 10/21/24, 2:18 AM Applications - Frictions and power transmission (October 11, 2024): Attempt review 2h 4a The distance between the CG of the figure and the point O is A1 + A2 (h + ) 3 3π 2h 4a A1 + A2 (h + ) 3 3π ȳ = = 0.25191598689683 m (*1) A1 + A2 The moment of inertia of the pendulum about pivot O is IO = 2.363193636835E − 5 kgm2 The horizontal projection of the motion of the any point on the pendulum can be described by x(t) = Xm ⋅ cos(2π 0.93290208504502 ⋅ t + ϕo ) The horizontal motion of the x(t) = Xm ⋅ cos(… t + ϕo ) = Xm ⋅ cos(5.8615966737921 ⋅ t + ϕo ) or 2π x(t) = Xm ⋅ cos(2π … t + ϕo ) = Xm ⋅ cos(2π ⋅ (0.93290208504502) ⋅ t + ϕo ) or x(t) = Xm ⋅ cos( … t + ϕo ) or Tp Tp 2 - - -Evaluate the length of the equivalent simple pendulum oscillating at the same frequency Leq = g( ) = 0.28552039988967 m 2π ω Calculate the speed of the motor Nmot = 60rev/min = 351.69580042753 rev/min 2π A gear box multiplies the speed of the motor by a factor 4 , if the wheel have a radius of 0.35 m, calculate the linear speed of the vehicle for a wheel of radius Rw : v = ω ⋅ n ⋅ Rw and the acceleration of the of the car a = 0 m/s2 --- Tp Calculate the distance travelled by the vehicle when the pendulum swings between two consecutive extreme postions ti = s 2 Tp The distance is therefore x = v and the speed when of the car when the pendulum is at its lowest point v = constant 2 https://moodlevut.vut.ac.za/mod/quiz/review.php?attempt=29842&cmid=464 8/8