Question

a salt shaker sits (0.102\text{ m}) from the center of a rotating lazy susan with (mu = 0.307).

what is the maximum speed of the salt shaker without sliding off?

? (\text{m/s})

Explanation:

🆕 New Concept Discovered: Centripetal Force and Static Friction
How friction provides the inward force for circular motion.

Step 1: Identify the forces acting on the salt shaker

For the salt shaker to move in a circle without sliding, the static friction force \( f_s \) must provide the necessary centripetal force \( F_c \).

The normal force \( F_N \) acting on the salt shaker on a flat surface is:

$$ F_N = mg $$

The maximum static friction force is:

$$ f_{s,\text{max}} = \mu F_N = \mu mg $$

Step 2: Set up the circular motion equation

The centripetal force required to keep the object moving in a circle of radius \( r \) at speed \( v \) is:

$$ F_c = \frac{mv^2}{r} $$

To prevent sliding, the required centripetal force cannot exceed the maximum static friction:

$$ F_c \le f_{s,\text{max}} $$

$$ \frac{mv^2}{r} \le \mu mg $$

Step 3: Solve for the maximum speed \( v \)

We can cancel the mass \( m \) from both sides of the inequality:

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

$$ v^2 \le \mu g r $$

$$ v \le \sqrt{\mu g r} $$

Thus, the maximum speed is:

$$ v_{\text{max}} = \sqrt{\mu g r} $$

Step 4: Calculate the numerical value

Substitute the given values into the formula:

• \( r = 0.102\text{ m} \)
• \( \mu = 0.307 \)
• \( g = 9.8\text{ m/s}^2 \) (standard acceleration due to gravity)

$$ v_{\text{max}} = \sqrt{0.307 \times 9.8 \times 0.102} $$

$$ v_{\text{max}} = \sqrt{0.3068732} $$

$$ v_{\text{max}} \approx 0.554\text{ m/s} $$

Answer:

0.554