Circular Motion Test. 2

Spacecraft are often placed in a stable, low parking orbit to check systems before moving to their final trajectory.

Since 2π radians = 360°, one radian = 360° / (2π) ≈ 57.3°.

The initial angle is arctan(3/2) ≈ 56.3°. The final angle is arctan(5/4) ≈ 51.3°. The angle decreased, so the displacement is clockwise.

The formula L = Iω is the rotational analogue of linear momentum p = mv.

The centripetal force is always perpendicular to the displacement. Since work is Fdcosθ and θ = 90°, the work done is zero.

By Newton's 3rd law, if A exerts a centripetal force on B, then B exerts an equal and opposite force back on A.

The minute hand completes one revolution (2π rad) in 60 minutes (3600 s). So, ω = 2π/3600 = π/1800 rad/s.

Torque is defined by a cross product, and its direction is determined by the right-hand rule, which is a convention.

The solid cylinder has a smaller moment of inertia, so more potential energy is converted into translational (linear) KE, making it faster.

Torque is force times the lever arm distance, so its unit is the product of the unit of force (Newton) and distance (meter).

Using the formula a_c = v²/r, we get (10 m/s)² / 50 m = 100 / 50 = 2 m/s².

This is derived from a_c = v²/r by substituting v = rω, which gives a_c = (rω)²/r = ω²r.

The total acceleration has a centripetal component (at 90° to velocity) and a tangential component (parallel to velocity). The resultant vector is at an acute angle.

This is a statement of Kepler's Third Law of planetary motion.

Since sitting on the disc is an internal process for the child-disc system with no external torque, total angular momentum is conserved.

The term 'uniform' in uniform circular motion specifically means that the magnitude of the velocity (the speed) is constant.

Just like the linear kinematic equations, the rotational versions assume that the angular acceleration is constant.

First, convert 540 rpm to rad/s: 540 × (2π/60) = 18π rad/s. Then, α = Δω/Δt = (18π - 0)/6 = 3π rad/s².

This is the standard formula for a thin rod rotated about its center of mass.

By pulling her arms in, she decreases her moment of inertia (I). Since angular momentum (L=Iω) is conserved, her angular speed (ω) must increase.