EX22.2 Flat Rolling Wheel

EX22.2 Flat Rolling Wheel - wheel to prevent slippage! What...

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A 50 lb wheel has a radius of gyration k G = 0.7 ft. Find the acceleration of the mass center if M = 35 lb-ft is applied. μ s = 0.3, μ k = 0.25. Rolling Kinetics. First step? ma Gx ma Gy I α G EOM x y F x = m( a G ) x F A = (50/32.2) a G F y = m( a G ) y N A – 50 = 0 M G = I G α - 35 + 1.25 F A = 0.761 α Note that I assumed everything is in the positive direction – I’ll let the algebra do the work. If the wheel is accelerating forward, the F will act to the right. Why? We assumed a was positive, which would make aG accelerate to the left, opposite our sign convention! N A = 50.0 lb, F A = 21.3 lb, α = -11.0 rad/s 2 , a G = 13.7 ft/s 2 3 equations, 4 unknowns. What next? Kinematics. First, assume the wheel is not slipping. Then, we can write a G = r (- α29 = -1.25 α Do the algebra. Need to check my no slip assumption Is F A = 21.3 lb μ s N A ? μ s N A =15 lbs, so can’t generate enough friction between the surface and the
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Unformatted text preview: wheel to prevent slippage! What do I need to do now? If slips, we know the friction force. What is it? F A = k N A = 0.25 N A (1) (2) (3) Now have 4 eq, 4 unknowns. Solve it Solving for the unknowns yields: N A = 50.0 lb F A = 0.25 N A = 12.5 lb = 25.5 rad/s 2 a G = 8.05 ft/s 2 (4) How would you find the acceleration of point A (touching the surface)? 2 / / G A G A G A r r a a A A A A A- + = Problem Type: Find: Given: k G = 0.7 ft, W = 50 lb, k = 0.25, s = 0.3, M = 35 ft-lbs a G Reflection: Need to be very careful with signs when considering rolling kinetics. Also when friction is involved you must check whether there is slip or not as two separate cases. This has implications for all types of vehicles from bicycles to race cars! EX22.2 Rolling Wheel on Flat Surface...
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This note was uploaded on 01/22/2011 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.

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