Question

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 vmax that the cylinder can have without slipping off the turntable?
7.3 m/s
0.73 m/s
0.25 m/s
0.53 m/s

Explanation:

Step1: Identify centripetal - friction force relationship

The maximum static - friction force provides the centripetal force for the cylinder not to slip. The maximum static - friction force is $F_f=\mu_sN$, and the centripetal force is $F_c = \frac{mv^2}{r}$. Also, $N = mg$ (in a vertical equilibrium situation, where the normal force equals the weight). So, $\mu_smg=\frac{mv^2}{r}$.

Step2: Solve for the maximum speed $v$

Cancel out the mass $m$ from both sides of the equation $\mu_smg=\frac{mv^2}{r}$, we get $\mu_sg=\frac{v^2}{r}$. Then, $v=\sqrt{\mu_sgr}$. Given $\mu_s = 0.45$, $g = 9.8\ m/s^2$, and $r=12\ cm=0.12\ m$.
Substitute the values: $v=\sqrt{0.45\times9.8\times0.12}=\sqrt{0.5292}\approx0.73\ m/s$.

Answer:

0.73 m/s