A uniform circular disk of mass 24.0 g and radius 40.0 cm hangs vertically from a fixed, frictionless, horizontal hinge at a point on its circumference. A horizontal beam of electromagnetic radiation with intensity 10.0 MW/m2 is incident on the disk in a direction perpendicular to its surface. The disk is perfectly absorbing, and the resulting radiation pressure makes the disk rotate. Find the angle through which the disk rotates as it reaches its new equilibrium position. (Assume that the radiation is always perpendicular to the surface of the disk.)
To find the angle through which the disk rotates, we need to calculate the torque experienced by the disk due to the radiation pressure and then use it to find the angular displacement.
First, let's find the torque. The torque is given by the formula τ = r × F, where r is the lever arm and F is the force creating the torque.
The force created by the radiation pressure is given by the formula F = P / c, where P is the power and c is the speed of light.
The power incident on the disk can be calculated by multiplying the intensity of the radiation by the area of the disk. The intensity of the radiation is given as 10.0 MW/m^2, and the area of the disk is πr^2, where r is the radius of the disk.
Let's plug in the values and calculate the power:
Area of the disk = π * (0.40 m)^2 = 0.16π m^2
Power = Intensity * Area = 10.0 MW/m^2 * 0.16π m^2 = 1.6π MW = 1.6π * 10^6 W
Now, let's calculate the force:
F = P / c = (1.6π * 10^6 W) / (3.0 * 10^8 m/s) = (1.6π / 3) N
Next, we need to calculate the lever arm, which is the radius of the disk. Given that the radius is 40.0 cm = 0.40 m, the lever arm is equal to the radius.
Now, let's calculate the torque:
τ = r × F = (0.40 m) × (1.6π / 3) N = 0.213 N · m
Finally, we can calculate the angular displacement. The torque τ is equal to the product of the moment of inertia I and the angular acceleration α. Since the disk is rotating from rest to equilibrium, we can assume the angular acceleration α is constant, and the torque τ is equal to I × α.
The moment of inertia I for a disk rotating about an axis passing through its center is given by the formula I = (1/2) m r^2, where m is the mass of the disk and r is the radius.
Let's plug in the values and find the angular displacement:
I = (1/2) m r^2 = (1/2) * (0.024 kg) * (0.40 m)^2 = 0.0048 kg · m^2
Since τ = I × α, we can rearrange the formula to find the angular acceleration α:
α = τ / I = (0.213 N · m) / (0.0048 kg · m^2) = 44.38 rad/s^2
Now, we can use the kinematic equation for angular displacement to find the angle through which the disk rotates. The equation is:
θ = (1/2) α t^2,
where θ is the angular displacement, α is the angular acceleration, and t is the time.
Since the disk is at rest initially, we can assume t = 1 s (time for it to reach equilibrium).
Let's calculate the angular displacement:
θ = (1/2) α t^2 = (1/2) * (44.38 rad/s^2) * (1 s)^2 = 22.19 radians
Therefore, the angle through which the disk rotates as it reaches its new equilibrium position is approximately 22.19 radians.