A uniform straight rod of mass M and length L is having small rollers of negligible mass at its ends. Rod is moving on smooth horizontal surface with constant speed V0 in gravity free space. Find the change in normal reaction at the rear roller from horizontal surface just after front roller crosses

Answer

Change in Normal Reaction on Rear Roller – Physics Explained

Change in Normal Reaction on Rear Roller When Front Roller Crosses an Edge

This is a conceptual and calculative mechanics problem involving a uniform rod with small rollers at both ends, moving over a smooth horizontal surface in gravity-free space. The question focuses on analyzing the change in normal forcefront roller crosses an edge (i.e., the front roller leaves the surface).

Understanding the Setup

Rod mass (M)
Rod length (L)
Constant speed = V₀ (no acceleration)
Gravity = 0 (space environment)
Surface is smooth ⇒ No frictional forces

Before Front Roller Crosses

Initially, both ends (rollers) are in contact with the surface. There is no vertical force from gravity, so the vertical reaction forces arise purely due to the **contact constraint** (maintaining the rod’s position along the surface).

    🔹 Both front and rear rollers apply equal and opposite normal reactions to keep the rod in horizontal motion. These are **internal constraints**.
  

After Front Roller Crosses the Edge

As soon as the front roller leaves the surface (e.g., the edge of a surface or platform), it can no longer exert a normal force. This leads to a **redistribution of support force** only through the rear roller.

Key Concepts

In gravity-free space, the rod is in **pure translational motion** at constant velocity.
Normal forces are **not balancing gravity**, but maintaining **constraints** (such as ensuring point contacts stay on surface).
Just after front roller leaves, the rod’s center of mass is no longer symmetrically supported.

Impulsive Effect and Rotation

At the instant the front roller crosses, an impulsive torque is generated due to the sudden **loss of front support**. This causes the rod to begin **rotating** about the rear roller, while still translating with speed V₀.

Analyzing Instant of Transition

    🔸 Since the front roller has just lost contact, the entire constraint (normal reaction) must now be provided by the rear roller.
  

Change in Normal Reaction = Final Reaction – Initial Reaction
Initial Reaction (rear) = 0.5 × M × a_normal
Final Reaction (rear) = Entire reaction = M × a_normal

But here, there is no gravity, and the surface is **horizontal and smooth**. The only acceleration in vertical direction is due to the change in contact constraint, not due to external vertical forces.

Result

    Change in Normal Reaction at Rear Roller = 0.5 M × V₀² / (L/2) = M × V₀² / L
  

This arises from the sudden initiation of rotation about the rear roller when the front roller loses contact.

Final Answer:

✅ Change in Normal Reaction = (M × V₀²) / L

Note: This is derived from the impulsive torque concept. Since gravity is absent, the reaction is not balancing weight but ensuring constraint of motion during transition.

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