A roller-caster vechicle has a mass of 500 kg when fully loaded with passengers (Fig P7.28) (a) If the vechile has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point? (b) What is the maximum speed the vehicle can have at point B in order for gravity to hold it on the track?(adsbygoogle = window.adsbygoogle || []).push({});

a) m = 500 kg

[tex] V_{t} = 20 m/s [/tex]

[tex] F_{t} = ? [/tex]

r = 10 m

[tex] a_{c} = \frac{v^2}{r} = 40 m/s^2 [/tex]

[tex] F = m( g + a_{c} ) = 2.49 * 10^4 N [/tex]

b) m = 500 kg

[tex] V_{t} = ? [/tex]

[tex] F_{t} = ? [/tex]

r = 15 m

[tex] F_{T} = n - F_{c} = m(g - a_{c} ) [/tex]

[tex] 0 = m(g - a_{c} ) [/tex]

[tex] g = a_{c} [/tex]

[tex] g = \frac{V_{t^2} }{r} [/tex]

[tex] V_{t} = \sqrt{9.8 m/s^2 * 15 m} = 12.1 m/s [/tex]

Could someone explain part (b). Why is there no acting F on the vehicle when it has reaced the maximum point on the roller coaster. Take a look at (Fig. P7.28).

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# Homework Help: Centripetal Acceleration of roller-caster vechicle

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