Question

In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25?

Answer 1

Answer:

Minimum speed will  be equal to 2.213 m/sec

Explanation:

We have given radius of the r = 2 m

Coefficient of friction $\mu =0.25$

At minimum speed frictional force will be equal to centripetal force

So $\mu mg=\frac{mv^2}{r}$

$\frac{v^2}{r}=\mu g$

$v=\sqrt{\mu rg}=\sqrt{0.25\times 2\times 9.8}=2.213m/sec$

So the minimum speed will be equal to 2.213 m/sec