Question

correct according to the discussion in example 7 in section 5.3, the maximum speed that the cylinder can have is given by $v_{max}=sqrt{mu_{s}gr}$, where $mu_{s}$ is the coefficient of static friction, $g$ is the acceleration due to gravity, and $r$ is the radius of the path. a small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. the cylinder is at a distance of $r = 12$ cm from the center of the turntable. the coefficient of static friction between the bottom of the cylinder and the surface of the turntable is 0.45. what is the maximum speed $v_{max}$ that the cylinder can have without slipping off the turntable?

Explanation:

Step1: Convert radius to SI - unit

We know $r = 12\ cm=0.12\ m$, $\mu_s = 0.45$, and $g = 9.8\ m/s^2$. The formula for the maximum speed is $v_{max}=\sqrt{\mu_sgr}$.

Step2: Substitute values into the formula

$v_{max}=\sqrt{0.45\times9.8\times0.12}$
$v_{max}=\sqrt{0.5292}$
$v_{max}\approx0.73\ m/s$

Answer:

$0.73\ m/s$