A block lies on a disk 0.2 m from the axis, μ = 0.5. At what angular velocity of the disk will it start to slip? — Physics problems | PhysLabo
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A block on a rotating disk: the limiting angular velocity

Medium 10 grade Mechanics / Circular Motion
Problem statement

A small block lies on a horizontal disk at a distance $r = 0.2$ m from the axis of rotation (see the figure). The coefficient of friction between the block and the disk is $\mu = 0.5$. The disk is spun up slowly. At what angular velocity $\omega$ will the block start to slip? Find the corresponding frequency of rotation $\nu$. Take $g = 10$ m/s².

Top view: a disk rotating with angular velocity ω and a small block at distance r from the axis
Given
  • $r = 0.2$ m
  • $\mu = 0.5$
  • $g = 10$ m/s²
  • $\omega = ?$, $\nu = ?$
Hints
Hint 1 / 2
Static friction gives the block its centripetal acceleration: $F_{fr} = m\omega^2 r$. Vertically $N = mg$.
Hint 2 / 2
Static friction is limited: $F_{fr} \le \mu N$. Slipping starts when $m\omega^2 r = \mu mg$ — the mass cancels.
Solution

While the block is at rest relative to the disk, it moves along a circle of radius $r$ with a centripetal acceleration $a = \omega^2 r$ directed toward the axis. Only static friction provides this acceleration; vertically the forces $m\vec g$ and $\vec N$ balance.

Side view: static friction provides the centripetal acceleration
Side view: static friction provides the centripetal acceleration

1. Vertically: $N = mg$. Horizontally (toward the axis): $F_{fr} = m\omega^2 r$.

2. Static friction cannot exceed $\mu N = \mu mg$. Slipping starts when it reaches its maximum:

$$m\omega^2 r = \mu mg \;\Rightarrow\; \omega = \sqrt{\frac{\mu g}{r}} = \sqrt{\frac{0.5 \cdot 10}{0.2}} = \sqrt{25} = 5 \text{ rad/s}$$

3. The frequency of rotation:

$$\nu = \frac{\omega}{2\pi} = \frac{5}{2 \cdot 3.14} \approx 0.80 \text{ Hz} \quad (\approx 48 \text{ rpm})$$
Answer

$\omega = 5$ rad/s; $\nu \approx 0.8$ Hz

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