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Fittoniya [83]
3 years ago
12

A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd

ing. The roadway is tilted toward the inside of the curve. Suppose the designated speed is v and the radius of the curve is R. At what angle should the curve be tilted?

Physics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:\theta =\tan ^{-1}(\frac{v^2}{gR})

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

R\cos \theta will balance weight and R\sin \theta will provide the necessary centripetal Force

thus R\sin \theta =\frac{mv^2}{R}  ------------1

R\cos \theta =m g  ----------------2

Divide 1 & 2 we get

\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}

\tan \theta =\frac{v^2}{gR}

\theta =\tan ^{-1}(\frac{v^2}{gR})                                      

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Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

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<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

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