JEE Physics Constrained Motion Notes Set 01

Constraints and Why They Matter

1. A constraint is a geometrical condition that links the motions of connected bodies, for example a taut string of fixed length, a rigid rod, or two bodies that stay in contact.
2. A constraint relation removes unknowns. Write it first, then differentiate once for velocities and twice for accelerations. Only after this combine with Newton's laws.
3. The relation holds as long as the string stays taut and inextensible, the rod stays rigid and the contact is not lost.

String Constraint: Fixed Pulley

1. Every point of a taut inextensible string moves along the string with the same speed.
2. If a body is attached to the string at an angle, only the component of that body's velocity along the string equals the string's speed. A body moving horizontally with speed v at a string making angle θ with the horizontal contributes \( v\cos\theta \).
3. For two bodies A and B tied to the same string over a fixed pulley, \( v_A\cos\theta_A = v_B\cos\theta_B \). For example if \( \theta_A = 53^\circ \), \( \theta_B = 37^\circ \) and \( v_A = 10 \) m/s then \( v_B = 10\times\frac{0.6}{0.8} = 7.5 \) m/s.
4. The same relation holds for accelerations when the geometry is fixed (angles not changing), but at an inclined string the angle changes with time, so accelerations need a separate derivation by differentiating the length relation.

Movable Pulley

1. If a pulley moves with velocity \( v_P \) and the two string ends attached to bodies move with \( v_A \) and \( v_B \) (all measured along the same line, with sign), then \( v_P = \frac{v_A+v_B}{2} \).
2. When one end is fixed, \( v_B = 0 \) and the free end moves at twice the pulley's speed: \( v_A = 2v_P \).
3. Equivalent form from the length of the string: for a rope with segments \( x_1, x_2, \dots \) hanging down, \( \sum x_i = \text{constant} \), where a movable pulley contributes twice its depth. Differentiating gives the relation between accelerations as well.
4. Use upward positive and substitute with signs. Example: pulley moving up at 10 m/s, block A moving down at 5 m/s gives \( 10 = \frac{-5+v_B}{2} \), so \( v_B = 25 \) m/s up.
5. A massless movable pulley is in equilibrium: the tension in the supporting string is twice the tension in the string passing over it.

Solving Complex Pulley Systems

1. Choose the longest continuous string that contains the point whose velocity is wanted.
2. Mark the points where the string touches real moving bodies. Parts touching fixed objects have zero velocity.
3. Take the rate of change of length of each part as positive if it increases the length and negative if it decreases it, then write: sum of all rates of change of length = 0.
4. Use the movable pulley relation to express velocities of the intermediate points, then solve the single equation for the unknown.

Rigid Rod Constraint

1. In a rigid rod or body the distance between any two points is constant, so the components of the velocities of any two points along the line joining them are equal.
2. Rod sliding with one end A on a vertical wall and the other end B on a horizontal floor, with angle θ between the rod and the floor: \( v_A\sin\theta = v_B\cos\theta \), so \( v_B = v_A\tan\theta \) in magnitude. As A moves down, B moves away from the wall.
3. At the instant when the rod is vertical the foot has zero horizontal speed while the top moves with the greatest speed.

Wedge Constraint

1. A body in contact with a wedge cannot move into it or away from it, so the components of velocity (and acceleration) of the two bodies perpendicular to the contact surface are equal.
2. A rod moving vertically with speed u on a wedge of angle θ moving horizontally with speed v: \( u\cos\theta = v\sin\theta \), so \( u = v\tan\theta \).
3. A block sliding down a wedge that itself moves has an acceleration that is the vector sum of its acceleration relative to the wedge (along the incline) and the wedge's acceleration.
4. Direction of contact force (normal reaction) is perpendicular to the contact surface; use it to write Newton's law along the common normal.

Moving-Pulley Block Systems Using Forces

1. After finding the constraint, apply Newton's second law to every body separately, with the same tension in a massless string over a frictionless pulley.
2. For a movable massless pulley the net force on it must be zero, giving the tension relation (2T on one side, T on the other for a single movable pulley).
3. Check the answer with the energy of the whole system, or by setting limiting cases where a body or the pulley is fixed.