Problem 6 - Master Resource Book in JEE Main Physics

Question

Two particles of equal mass are connected to a rope \(A B\) of negligible mass such that one is at end \(A\) and other dividing the length of rope in the ratio \(1: 2\) from \(B\). The rope is rotated about end \(B\) in a horizontal plane. Ratio of tensions in the smaller part to the other is (ignore the effect of gravity) (a) \(4: 3\) (b) \(1: 4\) (c) \(1: 2\) (d) \(1: 3\)

Step-by-Step Solution

Answer

The ratio of tensions is \(2:3\), marked by option (a).

Step 1: Understand the Setup

The problem involves two particles of equal mass connected to a rope that is rotating in a horizontal plane. One particle is at end \(A\) and the other divides the rope in a 1:2 ratio from end \(B\). We're asked to find the ratio of tensions in the smaller part (\(AB_2\)) to the longer part (\(B_2B\)).

Step 2: Determine Lengths of the Rope

Let's denote the total length of the rope as \(L\). Since the points divide the rope in the ratio 1:2 from \(B\), the length \(AB_2 = \frac{L}{3}\) while the length \(B_2B = \frac{2L}{3}\).

Step 3: Identify the Forces acting on the masses

The particle at end \(A\) only experiences the tension \(T_1\) since it's at the end of the rope. The particle at \(B_2\), in between, experiences two tensions: \(T_1\) pulling towards \(A\) and \(T_2\) pulling towards \(B\).

Step 4: Apply Centripetal Force Requirement at B2

Both particles rotate together with the rope, so at \(B_2\), the forces must provide the required centripetal force. Thus, tension \(T_2\) is greater than \(T_1\) to maintain circular motion of the section \(B_2B\). Mathematically, \(T_2 - T_1 = m\cdot \omega^2 \cdot \frac{L}{3}\) for the mass at \(B_2\).

Step 5: Express Tensions in terms of Rotation Rate

Similarly, for the particle at \(A\), the tension experienced \(T_1 = m \cdot \omega^2 \cdot \frac{2L}{3}\) because it experiences centripetal force due to its circular motion around \(B\) with a radius \(\frac{2L}{3}\).

Step 6: Find the Ratio of Tensions

Using both expressions from Step 4 and 5: \(T_2 = T_1 + m \cdot \omega^2 \cdot \frac{L}{3}\). Substituting \(T_1 = m \cdot \omega^2 \cdot \frac{2L}{3}\), we have \(T_2 = m \cdot \omega^2 \cdot \frac{3L}{3}\). Simplifying, get \(T_2 = m \cdot \omega^2 L\). Therefore, the ratio of \(T_1 : T_2 = 2: 3\).

Step 7: Calculate the Required Ratio

Since the tension we are asked for is \(T_1\) in the smaller section, divide by \(T_2\) in the larger part, giving \(T_1:T_2 = 2:3\). This implies the smaller part tension to larger is \(2:3\).

Key Concepts

Tension in Rope

When you think of tension in a rope during rotation, imagine pulling a string tied to a ball and spinning it in a circle above your head. The rope feels a pull, which is exactly what tension is — a force distributed along the rope. In the exercise, the rope is split into two segments. Each section feels different tensions because of the different lengths and positions of the masses.

• One mass is at the end of the rope, experiencing a straightforward tension since there’s no other mass beyond it to interact with.
• The second mass is not at the very end, so it feels a pull in both directions: toward the end and toward the rotating center.

This difference in tension between segments is what the problem emphasizes. Specifically, in ropes that form part of a rotating system, the inward section generally experiences greater tension since it pulls along the outer section too. The exercise provides the solution: the tension in the smaller section is 2 parts of the whole, while the larger section is 3 parts, reflecting the 2:3 tension ratio.

Rotation Dynamics

In rotation dynamics, objects moving in a circle constantly change direction, even if their speed is consistent. This change in direction indicates a form of acceleration called centripetal acceleration.

• For a rope rotating with particles at specific points, each part has to exert a force to keep the object moving in its circular path.
• This type of force, named centripetal force, always points toward the center of the circle.

The magnitude of the centripetal force required depends on factors such as the mass of the object, the radius of the circle, and the rotational speed (angular velocity, \(\omega\)). The formula used to calculate this force, \[F_c = m \cdot \omega^2 \cdot r\], illustrates how the force changes with the radius (distance from the center of rotation). In our exercise, as the rotation pivot is not at the rope's end but somewhere in between, the exercise highlights that different points along the rope need differing amounts of force to sustain rotation.

Mass Distribution

Understanding the mass distribution along a rope isn't just about the physical arrangement but also how it impacts forces and tensions in dynamic systems. The placement of mass influences the rotation much like how people on a merry-go-round feel different forces based on their seats (closer to or further from the center).

• In our scenario, two equal masses are deployed at different points on a rope.
• While these masses are identical, their positions cause differing effects on the rope's overall tension distribution when it's spun.

Keep in mind, the mass further from the rotation point (center) experiences more tension because it covers a larger circular path and needs more centripetal force to maintain its motion. Thus, in problems involving rotation, not only the amount but the distribution of mass requires careful consideration, as it shapes the tension experienced by different segments of the system.