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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose mag

Respuesta :

Answer:

The number of revolutions is 10.68 rev/min.

Explanation:

Given that,

Radius = 8 m

Suppose,  centripetal acceleration equal to the gravity

[tex]a_{c}=g=9.8[/tex]

We need to calculate the velocity

Using formula of centripetal acceleration

[tex]a_{c}=\dfrac{v^2}{r}[/tex]

[tex]v^2=a_{c}\times r[/tex]

Put the value into the formula

[tex]v=\sqrt{9.8\times8}[/tex]

[tex]v=8.85\ m/s[/tex]

We need to calculate the value of [tex]\omega[/tex]

Using formula of velocity

[tex]v=r\omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

Put the value into the formula

[tex]\omega=\dfrac{8.85}{8}[/tex]

[tex]\omega=1.12\rad/s[/tex]

We need to calculate the number of revolutions

Using formula of angular frequency

[tex]\omega=\dfrac{2\pi}{T}[/tex]

[tex]\omega=2\pi N[/tex]

[tex]N=\dfrac{\omega}{2\pi}[/tex]

Put the value into the formula

[tex]N=\dfrac{1.12}{2\pi}[/tex]

[tex]N=0.178\ rev/s[/tex]

Using conversion rev/s to rev/min

[tex]N=0.178\times 60[/tex]

[tex]N=10.68\ rev/min[/tex]

Hence,  The number of revolutions is 10.68 rev/min.