- #1

Samuelriesterer

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__Relative equations:__F = ma

a = v^2/r

Fcent = mv^2/r

ωf^2 – ωi^2 = 2αs

s = θr

ω = v/r

__Problem statement and work so far are all included:__(I am having trouble with 5 and no clue how to start on 7-9)

*You are designing a roadway for a local business, to include a circular exit ramp from the highway. The curve has a radius of 30 meters. The exit ramp is designed to function properly with no friction holding the cars on the ramp. To accomplish this, the ramp must be banked or sloped from one side to the other.*

**1)**If the design is to work for 1000 kg cars moving at 20 meters per second, calculate the force required to keep the car in its circular path.Fcent. = mv^2/r = (1000 kg)*(20 m/s)^2/(30 m) = 13333.3 N

**2)**Draw a force diagram of the car on the banked turn. Indicate which side is the outer radius of the turn.Only two forces, mg directed downward and N directed at an angle Θ left of the vertical axis.

**3)**Explain whether the normal force or the weight is larger and how you know (without calculation).Normal force because it is the force that is moving the car.

**4)**Calculate the angle the ramp should be sloped.Fnet_x = N sin θ = mv^2/r

Fnet_y = N cos θ = mg

N = mg/cos θ

mg/cos θ * sin θ = mv^2/r

mg *tan θ = mv^2/r

θ = arctan (20^2)/(30*9.8) = 53.68 deg

**5)**If the speed of a car on the ramp decreases from 20 m/s to 5 m/s after going around 3/4 of the 30 meter radius circle, determine the angular acceleration of the car.ωf^2 – ωi^2 = 2αs

s = θr = (3π/2)*30

ωi = v/r = 20/30 = 2/3

ωf = v/r = 5/30 = 1/6

α = (ωf^2 – ωi^2)/2s = [(2/3)^2 – (1/6)^2]/2*(3π/2)*30 = < I know this isn’t right>

**6)**If the coefficients of friction between tires and the road are 0.6 and 0.3 (you figure out which is which), determine the maximum speed a 1000 kg car can have and still make it around the curve. Start by redrawing the force diagram, including friction.N cos θ = mg + f sin θ

N = mg/(cos θ – μ sin θ)

Fnet = N sin θ + μ N cos θ = mg/(cos θ – μ sin θ) * (sin θ + μ cos θ) =

(mg tan θ + μ)/(1-μ tan θ) = mv^2/r →

v = sqrt [(rg tan θ + μ)/(1-μ tan θ)] = sqrt[(30*9.8 * tan 53.68 + .6)/(1-.6 * tan 53.68)] = 56.00 m/s

**7)**When the car is moving at 20 m/s, what is the speed of the center of the axle relative to the road??

**8)**Calculate the tangential speed (relative to the axle) of a point on the top of a 45 cm radius tire. explain your reasoning.**9)**Calculate the tangential speed (relative to the road) of a point on the top of a 45 cm radius tire.