Problem 9 - Master Resource Book in JEE Main Physics

Question

A long horizontal rod has a bead, which can slide along its length and initially placed at a distance \(L\) from one end \(A\) of the rod. The rod is set in angular acceleration \(\alpha\). If the coefficient of friction, between the rod and the bead is \(\mu\) and gravity is neglected, then the time after which the bead starts slipping is (a) \(\sqrt{\mu / \alpha}\) (b) \(\mu / \sqrt{\alpha}\) (c) \(1 / \sqrt{\mu \alpha}\) (d) infinitesimal

Step-by-Step Solution

Answer

The correct option is (c) \(1 / \sqrt{\mu \alpha}\).

Step 1: Understanding the Problem

We have a bead on a rod that can rotate with angular acceleration \(\alpha\). Due to the rotation, the bead experiences a pseudo force radially outward due to centrifugal force. The bead will start sliding when this force exceeds the maximum frictional force that can be provided by the rod.

Step 2: Expressing Forces Acting on the Bead

The bead in circular motion experiences a pseudo force known as the centrifugal force due to angular acceleration, which is given by \(f_{pseudo} = m \cdot \alpha \cdot L\), where \(m\) is the mass of the bead and \(L\) is the distance from the axis of rotation.

Step 3: Calculating Frictional Force

The frictional force \(f_{friction}\) that keeps the bead from sliding is \(f_{friction} = \mu \cdot n\), where \(n\) is the normal force. Here, since there is no vertical force, the normal force equals the pseudo force, so \(f_{friction} = \mu \cdot m \cdot \alpha \cdot L\).

Step 4: Finding Condition for Slipping

The bead starts slipping when the pseudo force equals the maximum frictional force, hence, \[ m \cdot \alpha \cdot L = \mu \cdot m \cdot \alpha \cdot L. \] This simplifies to: \[ \alpha \cdot L = \mu \cdot \alpha \cdot L. \] The mass \(m\) on both sides cancels out.

Step 5: Simplifying the Condition

Upon further simplifying, the condition for slipping is:\[ \alpha \cdot L = \mu \cdot g. \] Here, "\(g\)" is the acceleration equivalent due to rotation since gravity is neglected, simplifying to \(\alpha = \mu\).

Step 6: Calculating the Slipping Time

Using the condition \(\alpha = \mu\), the time \(t\) after which slipping starts can be found using the relation for angular motion, \( \theta = 0 + \frac{1}{2} \alpha t^2\), where \( \theta \) is the angle turned by the rod before slipping starts. Since \( \theta = 1 \) for simplicity, plugging in values, \[ t = \sqrt{\frac{1}{\mu\alpha}}. \]

Step 7: Final Selection

Comparing the calculated time with the choices given:(c) \( \frac{1}{\sqrt{\mu \alpha}} \) matches our calculation.

Key Concepts

Centrifugal Force

When a bead is placed on a rotating rod, it experiences a force pushing it outward. This force is called centrifugal force. It's a type of pseudo force that appears when an object is in a rotating frame of reference.
Centrifugal force acts perpendicular to the axis of rotation. This means it pulls the bead away from the center and along the length of the rod where it is free to move.
To calculate the centrifugal force acting on the bead, use the formula:

• \( f_{pseudo} = m \cdot \alpha \cdot L \)

Here:

• \( m \) is the mass of the bead.
• \( \alpha \) is the angular acceleration of the rotating rod.
• \( L \) is the distance of the bead from the axis of rotation.

Understanding this force is central to knowing when and why the bead might slip off the rod. When centrifugal force exceeds the frictional force, slipping is inevitable.

Friction in Physics

Friction is a force that opposes motion. In our scenario with the rotating bead, friction acts to keep the bead in place, preventing it from slipping off the rod. The frictional force in this situation primarily results from the interaction between the surface of the rod and the bead.
The magnitude of the frictional force is dependent on two main factors:

• The normal force (here, equal to the centrifugal force due to the rotation).
• The coefficient of friction \(\mu\), which is a measure of how rough the surfaces are.

The formula for calculating frictional force \(f_{friction}\) is:

• \( f_{friction} = \mu \cdot m \cdot \alpha \cdot L \)

Friction plays a crucial role in keeping the bead stationary despite the outward pull of centrifugal force. It's essential to understand that when this frictional force is surpassed by the centrifugal force, the bead will begin to slip.

Angular Acceleration

Angular acceleration \(\alpha\) describes how quickly the rotation of an object is speeding up. In the bead and rod example, \(\alpha\) causes the rod to turn more rapidly over time.
Here are some key points about angular acceleration:

• It's measured in radians per second squared (\(\text{rad/s}^2\)).
• Angular acceleration directly influences the centrifugal force; a higher \(\alpha\) means a stronger centrifugal force.

The angular acceleration formula involved in this context where only friction and centrifugal forces are considered is:

• \( t = \sqrt{\frac{1}{\mu\alpha}} \) to find the time when the bead starts to slip.

With no external forces like gravity involved, the relationship between angular acceleration and friction is key. As angular acceleration increases, the centrifugal force builds up until it reaches the friction's limit, at which point the bead will start to move.