Question #197185

1.Earth revolves around the sun with a speed of 30 km/s in an approximate circular path.

How much acceleration does it have directed towards the sun? Orbital radius of the earth

is 1.5 108 km.

2. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15o.

What is the radius of the loop?

3. A car of mass M experiences a maximum frictional force of 0.65Mg pavement as it goes

around the curved road of radius 150m. How fast can the car be moving to successfully go

around the curve?


1
Expert's answer
2021-05-27T18:37:59-0400

1) We can find the acceleration directed towards the Sun as follows:


ac=v2r=(3104 ms)21.51011 m=6.0103 ms2.a_c=\dfrac{v^2}{r}=\dfrac{(3\cdot10^4\ \dfrac{m}{s})^2}{1.5\cdot10^{11}\ m}=6.0\cdot10^{-3}\ \dfrac{m}{s^2}.


2) Let's apply Newton's Second Law of Motion in projections on axis xx- and yy:


Ncosθ=mg,Ncos\theta=mg,Nsinθ=mv2r.Nsin\theta=\dfrac{mv^2}{r}.

Dividing the second equation by the first one, we get:


tanθ=v2rg,tan\theta=\dfrac{v^2}{rg},r=v2gtanθ,r=\dfrac{v^2}{gtan\theta},r=(200 ms)29.8 mstan15=15.23 km.r=\dfrac{(200\ \dfrac{m}{s})^2}{9.8\ \dfrac{m}{s}\cdot tan15^{\circ}}=15.23\ km.

3)

Fc=Ffr,F_c=F_{fr},Mv2r=μMg,\dfrac{Mv^2}{r}=\mu Mg,v=μrg=0.65150 m9.8 ms2=31 ms.v=\sqrt{\mu rg}=\sqrt{0.65\cdot150\ m\cdot9.8\ \dfrac{m}{s^2}}=31\ \dfrac{m}{s}.

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