In an amusement park ride, a child stands against the wall of a cylindrical room that is then made to rotate. The floor drops downward and the child remains pinned against the wall. If the radius of the room is 2.15 m and the relevant coefficient of friction between the child and the wall is 0.600, with what minimum speed is the child moving if he is to remain pinned against the wall?

Respuesta :

Answer:

Minimum speed will be 2.7576 rad/sec        

Explanation:

We have given radius of the room r = 2.15 m

Coefficient of friction between the child and wall [tex]\mu =0.6[/tex]

Centripetal force acting on the child is given by [tex]F_C=\frac{mv^2}{r}[/tex], here m is mass r is radius and v is velocity

The normal force will balance the centripetal force

So [tex]N=\mu F_C[/tex]

[tex]\mu \frac{mv^2}{r}=mg[/tex]

[tex]v=\sqrt{\frac{rg}{\mu }}=\sqrt{\frac{2.15\times 9.81}{0.6}}=5.9289m/sec[/tex]

We know that [tex]v=r\omega[/tex]

So [tex]\omega =\frac{v}{r}=\frac{5.9289}{2.15}=2.7576rad/sec[/tex]