A transparent cube with index of refraction n = 1.6 and side length a = 4.0 cm is free to
rotate about an axis passing through its center. A laser beam is aimed at the cube. After passing through the cube the beam hits a screen on the other side forming a spot. The spot is displaced a distance y from the straight line path. The cube is then rotated at a constant angular velocity ω= 10 rad/s. As the cube rotates the spot moves. What is the speed of the spot at the instant that y = 0?
To find the speed of the spot at the instant when y = 0, we can use the concept of angular velocity and the relationship between linear and angular velocities.
First, let's analyze the situation. The laser beam travels through the transparent cube and gets refracted due to the change in medium. As the cube rotates, the refracted beam will change its direction, causing the spot on the screen to move.
To find the speed of the spot at the instant y = 0, we can break it down into two components: the linear velocity of the spot perpendicular to the laser beam's original path and the angular velocity of the cube.
Given:
Index of refraction, n = 1.6
Side length of the cube, a = 4.0 cm
Angular velocity, ω = 10 rad/s
To determine the linear velocity, we need to find the displacement of the spot per unit time, dy/dt.
Let's assume that the spot's displacement, y, is a function of time, t: y(t).
As the cube rotates, the displacement of the spot on the screen can be related to the rotation angle, θ, of the cube. The displacement y is given by:
y = a * tan(θ)
Differentiating both sides of the equation with respect to time, t, we get:
dy/dt = a * d(tan(θ))/dt
The derivative of the tangent function can be found using trigonometric identities. Since the cube is rotating at a constant angular velocity, we can express θ as a function of time:
θ = ω * t
Taking the derivative of θ with respect to time, we have:
dθ/dt = ω
Now, let's substitute the expression for θ back into the equation for dy/dt:
dy/dt = a * d(tan(θ))/dθ * dθ/dt
Since d(tan(θ))/dθ = sec^2(θ), we can substitute this value into the equation:
dy/dt = a * sec^2(θ) * dθ/dt
Substituting ω for dθ/dt, we have:
dy/dt = a * sec^2(θ) * ω
At the instant when y = 0, we know that tan(θ) = 0 because the spot is back on the straight line path. Therefore, θ = 0. Using the identity sec^2(0) = 1, we can simplify the equation:
dy/dt = a * 1 * ω
dy/dt = a * ω
Finally, to find the speed of the spot at the instant y = 0, we substitute y = 0 into the equation:
dy/dt = 0 * ω = 0
Therefore, the speed of the spot at the instant when y = 0 is zero.