How can I show that the sphere of radius R performs a simple harmonic movement. how can i set its reference point and make the f
ree body diagram.
I have the torque sum equation which is equal to the moment of inertia by angular acceleration
I have the torque sum equation which is equal to the moment of inertia by angular acceleration
1 answer:
Explanation:
Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod). There are three forces on the pendulum:
Weight force mg at the center of the sphere,
Reaction force in the x direction at the pivot,
Reaction force in the y direction at the pivot.
Sum the torques about the pivot O.
∑τ = I d²θ/dt²
mg (L sin θ) = I d²θ/dt²
For small θ, sin θ ≈ θ.
mg L θ = I d²θ/dt²
Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.
If you wish, you can use parallel axis theorem to find the moment of inertia about O:
I = Icm + md²
I = ⅖ mr² + mL²
mg L θ = (⅖ mr² + mL²) d²θ/dt²
gL θ = (⅖ r² + L²) d²θ/dt²
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Answer:
True
Explanation:
Answer:
Explanation:
<u>Given:</u>
- Current through the wire, passing through the origin,
- Current through the wire, passing through the y axis,
,
According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to times the net current threading the loop.
In case of a circular loop, the directions of magnetic field and the line element , both are along the tangent of the loop at that point, therefore,
.
is the circumference of the Amperian loop =
Therefore,
It is the magnetic field due to a current carrying wire at a distance r from it.
For the first wire, passing through the origin:
Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore,
The magnetic field at the given point due to this wire is given by:
For the first wire, passing through the y-axis:
Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore,
The magnetic field at the given point due to this wire is given by:
The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.
Thus, the net magnetic field at is given by