An electric dipole consists of 2.0 g spheres charged to 5.0 nC (positive and negative) at the ends of a 12 cm long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1400V, then released. What is the dipole’s angular velocity at the instant it is aligned with the electric field?
A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.42×105 , its speed is 27.7 , and the net braking force is , what is its speed 6.68 later?
To find the dipole's angular velocity at the instant it is aligned with the electric field, we can use the principle of conservation of energy.
First, let's calculate the torque on the dipole due to the electric field. The torque (τ) is given by the formula:
τ = pE sinθ
Where:
p = dipole moment (Product of magnitude of charge and distance between charges)
E = electric field strength
θ = angle between the dipole moment and the electric field vector
In this case, since the dipole is perpendicular to the electric field, θ = 90°, and sinθ = 1.
Now, let's calculate the dipole moment (p). The dipole moment (p) of an electric dipole is given by:
p = q * d
Where:
q = magnitude of charge on one end of the dipole
d = distance between the charges
In this case, the magnitude of charge on one end of the dipole (q) is 5.0 nC, and the distance between the charges (d) is 12 cm (or 0.12 m).
p = (5.0 x 10^-9 C) * (0.12 m) = 6.0 x 10^-10 C.m
Now, we can calculate the torque (τ) using the formula:
τ = pE
Substituting the values:
τ = (6.0 x 10^-10 C.m) * (1400 V)
τ = 8.4 x 10^-7 N.m
Since there are no external torques acting on the dipole (due to the frictionless pivot), the dipole's initial mechanical energy is equal to the final mechanical energy when it is aligned with the electric field.
The mechanical energy (E) of the dipole is given by:
E = (1/2) * I * ω^2
Where:
I = moment of inertia of the dipole
ω = angular velocity of the dipole
The moment of inertia (I) for a point mass rotating about an axis perpendicular to it is given by:
I = m * r^2
Where:
m = mass of each sphere
r = distance from the pivot point to the center of the sphere
In this case, the mass of each sphere (m) is 2.0 g, which is 0.002 kg. The distance from the pivot point to the center of the sphere (r) is half of the length of the rod, which is 6 cm (or 0.06 m).
I = (0.002 kg) * (0.06 m)^2 = 7.2 x 10^-7 kg.m^2
Now, let's substitute the values into the expression for mechanical energy:
E = (1/2) * (7.2 x 10^-7 kg.m^2) * ω^2
At the instant the dipole is aligned with the electric field, the torque does work on the dipole to change its energy from potential energy to kinetic energy.
Thus, the initial potential energy of the dipole is equal to the final kinetic energy:
τ * θ = (1/2) * I * ω^2
Since θ = 90° (aligning the dipole with the electric field), the left-hand side of the equation becomes τ * sinθ = |τ|.
|τ| = (1/2) * I * ω^2
Now we can solve for the angular velocity (ω):
ω^2 = (2 * |τ|) / I
Therefore:
ω = √((2 * |τ|) / I)
Plugging in the values:
ω = √((2 * (8.4 x 10^-7 N.m)) / (7.2 x 10^-7 kg.m^2))
ω ≈ 14.14 rad/s
So, the dipole's angular velocity at the instant it is aligned with the electric field is approximately 14.14 rad/s.