A thin, uniform rod with negligible mass and length 0.200 m is attached to the floor by a frict

A thin, uniform rod with negligible mass and length 0.200 m ...

A thin, uniform rod with negligible mass and length 0.200$\mathrm { m }$ is attached to the floor by a frictionless hinge at point $P$ (Fig. P27.79). A horizontal spring with force constant $k = 4.80 \mathrm { N } / \mathrm { m }$ connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field $B = 0.340$ T directed into the plane of the figure. There is current $I = 6.50 \mathrm { A }$ in the rod, in the direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at $P .$ Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of $53.0 ^ { \circ }$ with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium?

A thin, uniform rod with negligible mass and length
0.200$\mathrm { m }$ is attached to the floor by a
frictionless hinge at point $P$ (Fig.
P27.79). A horizontal spring with
force constant $k = 4.80 \mathrm { N } / \mathrm { m }$ connects the other end of the rod to a vertical wall. The rod is in a uniform
magnetic field $B = 0.340$ T directed
into the plane of the figure. There is
current $I = 6.50 \mathrm { A }$ in the rod, in the
direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at $P .$ Is it
correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of $53.0 ^ { \circ }$ with the floor, is
the spring stretched or compressed? (c) How much energy is stored
in the spring when the rod is in equilibrium?

Key Concepts

Magnetic Force on Current-Carrying Conductors

This concept involves the force experienced by a current-carrying conductor in a magnetic field. The force is given by the cross product F = I (L × B), where I is the current, L is the length vector of the conductor, and B is the magnetic field. Understanding this principle is key in analyzing devices that rely on electromagnetic interactions, such as motors and generators.

Torque

Torque is a measure of the tendency of a force to produce rotational motion about an axis. It is calculated as the product of the force and the perpendicular distance (lever arm) from the pivot point. This concept is fundamental in rotational dynamics and is used to determine how forces affect the angular motion of rigid bodies.

Center of Force Application

This concept addresses how distributed forces can be represented by a single equivalent force acting at a specific point, often the center of mass or center of gravity. However, caution must be exercised since this simplification is valid for gravitational forces in uniform fields but may not hold when dealing with other forces like magnetic forces, where the distribution of the force over the length of an object can impact the resulting torque differently.

Equilibrium Conditions in Statics

Equilibrium in statics requires that the net force and net torque acting on a system are zero. This concept is crucial for solving problems where objects are at rest, ensuring that all forces and moments balance. It is widely used in engineering and physics to analyze structures and mechanical systems.

Elastic Potential Energy in Springs

This concept refers to the energy stored in a spring when it is either stretched or compressed. It is quantified by the expression U = ½ k x², where k is the spring constant and x is the displacement from its equilibrium position. This energy becomes important when analyzing systems involving elastic components, as it contributes to the overall energy balance of the system.

Transcript

00:01 In the given problem, there is a thin, uniform rod with a negligible mass, which is attached the spring to a hinge point p, as shown here in the figure.

00:29 This is the rod.

00:36 One of its end is fixed here at the point p, and its another end is attached with a point p, and its another a string this point p is frictionless and there is a uniform magnetic field which is coming out perpendicular which is getting in to the plane of paper perpendicularly the magnitude of this magnetic field is given as 0 .340 tesla there is a current passing through this rod in downward direction whose value is given as i is equal to 6 .50 ampere the total length of this rod is given as l is equal to 0 .2200 meter the spring factor for this spring the given spring is k is equal to 4 .80 newton per meter.

02:03 Now in the presence of electric current passing through this rod and the magnetic field, this roar will start experiencing a magnetic force in the direction as shown here.

02:19 And this force, the expression for this force is given as is equal to b into i into l and using clemmings left -hand rule the direction of this force is as shown in the figure now in the first part of the problem we have to find the torque due to this magnetic force on the rod so to find this torque we assume this force to be acting at the center of mass of the rod if the force is a assumed to be acting at the center of mass of the rod as shown in the figure, then the torque acting on this rod will be given by tau is equal to the value of this force, which is f multiplied by the perpendicular distance of this force from the point of rotation means p and that is equal to half of the length so we can say it becomes b i l multiplied by l by two or simply b i l square by so when we plug in all the known values for magnetic field this is 0 .340 tesla multiplied by the current which is 6 .50 ampere the length, the square of 0 .200 and divided by that is in meter...

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