A horizontal bar of mass m1 and small diameter is suspended...

A horizontal bar of mass $m_{1}$ and small diameter is suspended by two wires of length $l$ from a carriage of mass $m_{2}$ which is free to roll along the horizontal rails. If the bar and carriage are released from rest with the wires making an angle $\theta$ with the vertical, determine the velocity $v_{\text {ble }}$ of the bar relative to the carriage and the velocity $v_{c}$ of the carriage at the instant when $\theta=0 .$ Neglect all friction and treat the carriage and the bar as particles in the vertical plane of motion.

Key Concepts

Relative Motion

Relative motion involves analyzing the velocity of one object with respect to another. It is a critical concept when different parts of a system move with different speeds and directions. In scenarios where a component is moving relative to another body (for instance, a bar moving relative to a carriage), understanding how to superimpose or subtract velocities is crucial to accurately describe the system's dynamics.

Conservation of Energy

This principle states that in a closed system with no non?conservative forces like friction, the total mechanical energy remains constant. In dynamics problems, it allows one to relate the loss of gravitational potential energy to the gain in kinetic energy of the system’s components. In such problems, identifying the conversion from potential to kinetic energy is fundamental for solving for velocities at different configurations.

Conservation of Momentum

With no external forces acting along a particular direction, the total momentum in that direction is conserved. In problems with moving systems like a rolling carriage, conservation of momentum, particularly in the horizontal direction, is essential for relating the individual momenta of different objects within the system, ensuring that the center of mass motion is correctly accounted for.

Constraint Relationships

Constraint relationships refer to the fixed geometrical or physical connections between components in a mechanical system. For systems where lengths (such as wires or rods) remain constant, these constraints create equations that relate the motions of different parts of the system. Utilizing these constraints enables a reduction in the number of independent variables that must be solved for in the problem.

Transcript

00:01 Hi, in the given problem.

00:03 Suppose this is the pulley, the massless and frictionless pulley.

00:09 This is the cord.

00:13 Here, this is the cross section of the beam, which is moving perpendicular to this magnetic field.

00:21 Here, these are the magnetic field lines.

00:25 The mass of this rod is m.

00:28 Then here, this is the block.

00:31 It vertically from this string having the mass capital m.

00:37 So the weight of this block, capital mg, acting vertically downward, tension in this string, and here this is also the tension in the string.

00:49 Now if we consider the free body diagram of this suspended block as the block is moving down.

00:56 Suppose it is moving down with an acceleration a.

00:58 So the force equation will be the net force acting on it will be m g minus t, which using newton's second law of motion will be equal to m .a.

01:11 Then if you consider the motion of this beam perpendicular to the magnetic field, an emf will be induced.

01:19 This is equation number one.

01:21 An emf will be induced.

01:22 Emotionally will be induced across its ends given by bvl, where this v is the incidence.

01:29 Is tantaneous speed of this beam perpendicular to the magnetic field.

01:34 L is the length of this beam.

01:36 So the current passing through this beam will be emf by r using holmes law.

01:44 So this is bvl by r.

01:47 Hence the force, magnetic force acting on this beam opposite in direction to the motion of this beam as per lens's law will be given as b, i, hence it becomes b into b v l divided by r into l so finally this is b l to the whole square into v r so if i use now the free body diagram of this beam this was the f bd of suspended block and here now we are using f, b, d, free body diagram of the beam within the magnetic field.

02:43 So as the beam is moving towards left and it will be moving with the same acceleration a with which this block will be falling down.

02:52 So this is t minus f is equal to small m into a or we can say this is t minus b l to the whole square into v by r is equal.

03:06 To m .a and this is equation number two.

03:09 Now adding equation 1 and 2 m g minus t the left -hand side of equation 1 and t minus b l square v by r b l to the whole square v by r the left -hand side of the equation number 2 is equal to m a right -hand side of first equation is equal to small m a right -hand of second equation...

Please give Ace some feedback