Two blocks of masses m1 = 10 kg and m2 = 5 kg connected to each other by a massless inextensible str

Q: Why does friction act inwards on $m_1$?

The required centripetal force for $m_1$ ($124\text{ N}$) is greater than the tension provided by the string ($88\text{ N}$). To make up for this deficit and keep $m_1$ in circular motion, static friction must act in the same direction as tension, which is towards the center.

Q: Why is the tension the same for both blocks?

The string is given as massless and inextensible, and the connection is ideal. In such ideal strings, tension is uniform throughout its length.

Q: What would happen if $m_2$ also had friction?

If $m_2$ had friction, the tension wouldn't be uniquely determined just by its centripetal requirement. We would have a range of possible tensions depending on how static friction distributes itself between the two blocks to prevent slipping.

Q: Why do we use static friction and not kinetic friction?

The problem states the masses are "at rest with respect to an observer on the turn table." This means they are not sliding across the table's surface. Therefore, the friction acting on them is static friction, which adjusts its value to maintain this relative rest up to its maximum limit.

Q: In part (c), why does the angular velocity $\omega$ cancel out?

When we set friction to zero, the tension must perfectly provide the centripetal force for both masses: $T = m_1 r_1 \omega^2$ and $T = m_2 r_2 \omega^2$. Equating them gives $m_1 r_1 \omega^2 = m_2 r_2 \omega^2$. Since $\omega$ is non-zero, we can divide both sides by $\omega^2$, showing that the zero-friction position is independent of how fast the table is spinning!

Question

Two blocks of masses $m_1 = 10 	ext{ kg}$ and $m_2 = 5 	ext{ kg}$ connected to each other by a massless inextensible string of length $0.3 	ext{ m}$ are placed along a diameter of turn table. The coefficient of friction between the table and $m_1$ is $0.5$ while there is no friction between $m_2$ and the table. The table is rotating with an angular velocity of $10 	ext{ rad/s}$ about a vertical axis passing through its centre $O$. The masses are placed along the diameter of table on either side of the centre $O$ such that the mass $m_1$ is at a distance of $0.124 	ext{ m}$ from $O$. The masses are observed to be at rest with respect to an observer on the turn table.
(a) Calculate the frictional force on $m_1$.
(b) What should be the minimum angular speed of the turn table, so that the masses will slip from this position?
(c) How should the masses be placed with the string remaining taut so that there is no frictional force acting on the mass $m_1$?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a dynamic scenario: a turntable spinning with an angular velocity $\omega$, carrying two masses, $m_1$ and $m_2$, connected by a taut, massless string. The masses are positioned on opposite sides of the center $O$. The table is rough for $m_1$ (coefficient of friction $\mu = 0.5$) but perfectly smooth for $m_2$.

Our goal is to understand the delicate interplay of forces keeping these masses in uniform circular motion without slipping.

### The Master Equation for $m_2$

Let's start with the simpler mass, $m_2$. Since there is no friction between $m_2$ and the table, the only horizontal force acting on it is the tension $T$ from the string. For $m_2$ to maintain its circular path of radius $r_2$, this tension must provide the exact centripetal force required.

$$T = m_2 r_2 \omega^2$$

We know the total length of the string is $l = 0.3	ext{ m}$, and $m_1$ is at $r_1 = 0.124	ext{ m}$. Thus, we can easily find $r_2$:

$$r_2 = l - r_1 = 0.3 - 0.124 = 0.176	ext{ m}$$

Substituting the given values ($m_2 = 5	ext{ kg}$, $\omega = 10	ext{ rad/s}$):

$$T = 5 	imes 0.176 	imes (10)^2 = 88	ext{ N}$$

The string is under a tension of $88	ext{ N}$.

### The Friction on $m_1$

Now, let's shift our focus to $m_1$. It also needs a centripetal force to stay in its circular orbit of radius $r_1$. Let's calculate how much force it actually needs:

$$F_{c1} = m_1 r_1 \omega^2 = 10 	imes 0.124 	imes (10)^2 = 124	ext{ N}$$

Here lies the crux of the problem! Mass $m_1$ requires $124	ext{ N}$ pulling it towards the center, but the string is only pulling it with a tension of $88	ext{ N}$. Where does the missing force come from?

This is where static friction steps in as the hero. To prevent $m_1$ from sliding outwards, static friction $f$ must act **inwards** (towards the center) to assist the tension.

$$f + T = F_{c1}$$

$$f = 124 - 88 = 36	ext{ N}$$

So, the frictional force acting on $m_1$ is **$36	ext{ N}$ towards the center**.

### Pushing the Limits: When Does it Slip?

As the turntable spins faster (increasing $\omega$), the required centripetal force increases, and consequently, the required friction increases. However, static friction has a maximum limit, $f_{\max}$.

$$f_{\max} = \mu m_1 g = 0.5 	imes 10 	imes 9.8 = 49	ext{ N}$$

Slipping will commence the moment the required friction exceeds this maximum limit. Let's write the general expression for the required friction:

$$f = m_1 r_1 \omega^2 - m_2 r_2 \omega^2 = (m_1 r_1 - m_2 r_2)\omega^2$$

To find the minimum angular speed for slipping, we set this required friction equal to $f_{\max}$:

$$(10 	imes 0.124 - 5 	imes 0.176)\omega^2 = 49$$

$$(1.24 - 0.88)\omega^2 = 49$$

$$0.36 \omega^2 = 49$$

$$\omega = \sqrt{\frac{49}{0.36}} = \frac{7}{0.6} \approx 11.67	ext{ rad/s}$$

The masses will start slipping when the angular velocity reaches **$11.67	ext{ rad/s}$**.

### The Perfect Balance: Zero Friction

Finally, consider a fascinating hypothetical: how should we place the masses so that $m_1$ requires absolutely zero friction to stay in place?

If $f = 0$, it means the tension alone is perfectly providing the centripetal force for both masses simultaneously.

$$m_1 r_1 \omega^2 = m_2 r_2 \omega^2$$

Notice how beautifully $\omega^2$ cancels out! This means the zero-friction position is independent of the turntable's speed.

$$m_1 r_1 = m_2 r_2$$

$$10 r_1 = 5 r_2 \implies r_2 = 2 r_1$$

We also have the geometric constraint that the total length of the string is $0.3	ext{ m}$:

$$r_1 + r_2 = 0.3$$

Substituting $r_2 = 2 r_1$ into this equation:

$$r_1 + 2 r_1 = 0.3$$

$$3 r_1 = 0.3 \implies r_1 = 0.1	ext{ m}$$

To achieve a state of zero friction, mass $m_1$ must be placed exactly **$0.1	ext{ m}$** from the center.

Visualized Solution

$Visualizing the Setup$

$Dynamics of m_{2}$

$Calculating r_{2}$

$Evaluating Tension T$

$Dynamics of m_{1}$

$Direction of Friction$

$Calculating Friction f$

$Limiting Friction$

$Condition for Slipping$

$Calculating ω_{m i n}$

$Zero Friction Condition$

$Relating r_{1} and r_{2}$

$Final Position of m_{1}$

The Sigma Insight: Dynamics of Circular Motion

Analyzing the Setup

The Master Equation for $m_{2}$

The Friction on $m_{1}$

Pushing the Limits

The Perfect Balance

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dynamics of m2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating r2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Tension T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dynamics of m1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction of Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Friction f

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Slipping

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ωmin​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zero Friction Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating r1​ and r2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Position of m1​