7 A block of mass m₁ kg rests on a horizontal turntable rotating uniformly at o rad/s. A string attacted to thin block passes through a hole in the centre of the table and supports another block of mass m₂ kg. The coefficient of static friction between the first block and the table is µ. Find the ratio of the maximum and minimum values of R for which the first block does not slide on the turn table [There is no friction between string & table

Concept-by-Concept Walk-through

1. Uniform Circular Motion
   A body of mass $m_1$ moving in a circle of radius $R$ at angular speed $\omega$ needs a centripetal force

   $F_c = m_1 \omega^2 R.$

2. Tension in the String
   The string passes through a smooth (friction-less) hole. The hanging block $m_2$ is at rest, so it obeys

   $T = m_2 g.$

3. Static Friction on the Top Block
   Static friction can act either way (inward or outward). Its magnitude can be anything from $0$ up to

   $f_{\text{max}} = \mu m_1 g.$

4. Force Balance along the Radial Direction
   Radially inward is taken as positive. The net radial force on $m_1$ must equal the required centripetal force:

   $T \;\pm\; f = m_1 \omega^2 R.$

   • Use +f when friction points inward.
   • Use −f when friction points outward.

5. Finding the Extreme Cases
   Sliding starts when $|T - m_1 \omega^2 R|$ just equals $f_{\text{max}}$. Therefore

   $|\,m_1 \omega^2 R - T\,| \le \mu m_1 g.$

   Substitute $T = m_2 g$ and split the absolute-value into two inequalities to get the smallest ($R_{\min}$) and largest ($R_{\max}$) allowed radii.

6. Common Pitfalls
   • Remember friction’s direction depends on which way sliding would begin.
   • Ensure $m_2 > \mu m_1$—otherwise the lower block cannot even lift the top block’s frictional limit, and only one extreme exists.

What is happening here?

Imagine you are spinning a plastic lazy-Susan (the round tray) and you place a toy block on it. The block is tied with a string that goes through a tiny hole at the centre and hangs another block below the table.

• The top block wants to fly outward because the table is spinning.
• The hanging block wants to pull the string downward because of gravity.
• Friction between the top block and the spinning plate tries to keep the top block from sliding.

We must find the biggest and smallest distances $R$ (how far the top block sits from the centre) for which friction can still prevent sliding. Then we take the ratio of these two distances.

The game is about balancing three forces:

1. Centripetal pull needed to move in a circle.
2. Tension in the string (because of the lower block).
3. Static friction that can act either inward or outward (whichever is needed) but only up to a maximum strength given by $\mu m_1 g$.

If the required balance is just at the limit of friction, we get the extreme values of $R$.

Step-by-Step Calculation

1. Forces on the Top Block ($m_1$)

   • Inward tension: $T = m_2 g$
   • Static friction: $f$ (direction will depend)
   • Required centripetal force: $m_1 \omega^2 R$

2. Inequality for No-Sliding Condition

   $|m_1 \omega^2 R - T| \le \mu m_1 g.$

3. Insert $T = m_2 g$

   $|m_1 \omega^2 R - m_2 g| \le \mu m_1 g.$

4. Split the Absolute Value

   (i) $m_1 \omega^2 R - m_2 g = \mu m_1 g$  →  extreme outward friction (gives $R_{\max}$)
   (ii) $m_2 g - m_1 \omega^2 R = \mu m_1 g$  →  extreme inward friction (gives $R_{\min}$)

5. Solve for $R_{\max}$

   $m_1 \omega^2 R_{\max} = m_2 g + \mu m_1 g$

   $\Rightarrow \boxed{\;R_{\max} = \frac{g\,(m_2 + \mu m_1)}{m_1 \omega^2}\;}$

6. Solve for $R_{\min}$

   $m_1 \omega^2 R_{\min} = m_2 g - \mu m_1 g$

   $\Rightarrow \boxed{\;R_{\min} = \frac{g\,(m_2 - \mu m_1)}{m_1 \omega^2}\;}$

   (Valid only if $m_2 > \mu m_1$.)

7. Required Ratio

   $\frac{R_{\max}}{R_{\min}} = \frac{m_2 + \mu m_1}{m_2 - \mu m_1}.$

8. Final Answer

   $\boxed{\displaystyle \frac{R_{\max}}{R_{\min}} = \frac{m_2 + \mu m_1}{m_2 - \mu m_1}}$