Physics 1301 - University of Minnesota

Physics 1301University of MinnesotaM. ZudovYou are designing a banked curve for a new section of interstate 35. The specificationsyou are given are:1. During icy conditions in the winter, when the coefficient of static friction is 0.15, a carwhich stops on the curve must not slide down into the ditch at the edge of the road.2. In the summer, when the coefficient of static friction is 0.4, a car must be able to roundthe curve without skidding at a maximum speed of 60 miles/hour.3. The radius of the curve must be as small as possible.(a) What is the design value for the angle at which the curve should be banked?(b) What is the design value for the radius of the curve?(c) In winter what is the maximum speed for a car to round the curve without skidding?(d) What should be the speed of a careful motorist who does not want to wear her tires byusing the force of friction?Bonus:Assume that the road is not banked.(e) What would be the maximum speed in winter (same radius)?(f) What would be the radius of the curve to satisfy condition 2. above?μ = 0.15, μ = 0.40, vWy : masinθ= N − mg cosθ⇒ N = masinθ+ mg cosθx : ma cosθ= mg sinθ− μN= mg sinθ− μmasinθ− μmgcosθ⇒2va =RμW− μ(a) v = 0, μ = μW⇒ tanθ= μW⇒ θ = 8.53° ⇒ a = g1+μμ(b) v = v(d) μ = 0 ⇒ v(e) θ = 0 ⇒ vsinθ− μ cosθtanθ− μ=g = gcosθ+ μ sinθ1+μ tanθ(c) μ = −μ⇒ vS(f) θ = 0, v = v21−μStanθvS, μ = −μS⇒ R =tanθ+ μ gWS22S2μW+ μW=1+μ μμW− 0=1+0⋅μ= 60 mph = 26.8 m/s, f = μN2μWRg =21−μ1−μSμW=μ + μ0 + μW= gR = μWgR = ...(same as above)1−μ ⋅0WSWW2 21−μS⋅0vSvS⇒ R ==0 + μ g μ gW1−μSμWvμ + μ= 183.5 m2S1−μSμWgR = μWgR = μWvμ + μSSSWWWWSSS2vS= 125.5 mg⇒ v = v2SWS⇒ v = vS2μW21−μW1−μSμWμ + μ= 0.72⋅vμW(1 − μSμW)= 0.50⋅vSμ + μWWSSS= 43.5 mph= 30.4 mph

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Physics 1301University of MinnesotaM. Zudov1. Block (mass m 2 ) is placed on top of a prism (mass m 1 , angle α ). The coefficient offriction between the block and the prism is μ. Find the maximum and minimumhorizontal accelerations of the prism at which the block will not slide.(a) θ = 10°, μ = 0.4, aa − ?yx22: m a sinα= N − m: m a cosα= msinα± μ cosα± μ + tanαa = g= gcosαm μ sinα1mμ tanαaamaxmin222μ + tanα= g = 0.62g1−μ tanα− μ + tanα= g = −0.21g1+μ tanα2g cosα⇒ N = mg cosα+ m a sinαm2acosα− mg sinα± μN⇒ N =± μ22= a = a± μmg cosα± μm a sinα= m a cosα− m212222g sinα⇒2g sinα

Physics 1301University of MinnesotaM. ZudovBlock (mass m 2 ) is placed on top of a prism (mass m 1 , angle α ).Assuming that friction is negligible, find the acceleration of the prism, a 1 .(a) θ,m , m211- ?2 y211: ma2 y211y22We assume xa, μ = 01to the right, ySince there is no relative motion along yay= ax : m a= −asinα= N sinα= m21122⊥,up the incline= −ma sinα= N − m g cosα⇒ N = mg cosαsinα− m a22, both the block and the prizm have the same acceleration projections on y2122sin α ⇒ ag cosα− m112a sinαm2cosαsinα= g2m + m sin α122.

Physics 1301University of MinnesotaM. ZudovA block starts sliding off the top of a wedge whose base is equal to l = 2.10 m. Thecoefficient of friction between the block and the wedge is equal to μ = 0.140. What angleα corresponds to a minimum sliding time? How would this angle change if thecoefficient of friction is set to zero?m = 2.00 kg, l = 2.10 m, μ = 0.140, t = tα − ?a = g(sinα− μ cosα)min, v02l at 2 2l2lx = = ⇒ t = =cosα2 a cosαg cosα(sinα− μ cosα)dcosα(sinα− μ cosα)= −sinα(sinα− μ cosα)+ cosα(cosα+ μ sinα)=dα221= cos α − sin α + 2μcosαsinα= cos 2α+ μ sin 2α= 0 ⇒ tan 2α= − < 0 ⇒ 2α> 90°μ1 ⎛ ⎞=−11α tan ⎜ − ⎟ = 98°⇒ α = 49°2 ⎝ μ ⎠μ = 0 ⇒ α = 45°= 0

Physics 1301University of MinnesotaM. ZudovA small block A of mass m =5 kg is launched off the top of a sphere of radius R = 1 m.The block breaks off the sphere with velocity v = 2 m/s. The coefficient of frictionbetween the block and the sphere is equal to μ = 0.15. Find the angle θ corresponding tothe point at which the block breaks off the sphere.m,R,v,μθ − ?may2v= mR(break - off= mg cosθ− Npoint) ⇒ N= 0 ⇒2vR= g cosθ⇒ θ = cos−1( v2/ gR)= 65.9°