Question

current attempt in progress
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?
0.25 m/s
0.53 m/s
0.73 m/s
7.3 m/s

Explanation:

Step1: Identify centripetal force

The maximum static - friction force provides the centripetal force for the cylinder not to slip. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, and the maximum static - friction force is $F_{s,max}=\mu_s N$. On a horizontal surface, $N = mg$. So $F_{s,max}=\mu_smg$.

Step2: Equate forces

Set $F_c = F_{s,max}$, we have $\frac{mv^2}{r}=\mu_smg$. The mass $m$ cancels out on both sides of the equation, and we get $v^2=\mu_sgr$.

Step3: Convert units

Given $r = 12\ cm=0.12\ m$, $\mu_s = 0.45$, and $g = 9.8\ m/s^2$.

Step4: Calculate maximum speed

Substitute the values into the formula $v=\sqrt{\mu_sgr}=\sqrt{0.45\times9.8\times0.12}=\sqrt{0.5292}\approx0.73\ m/s$.

Answer:

0.73 m/s