Important Rotational And Circular Motion Quiz Test Online

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.

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

Tension has a vertical component that balances weight and a horizontal component that provides the centripetal force.

Moment of inertia I ∝ R². If R becomes R/2, I' becomes I/4. By conservation of angular momentum (Iω = I'ω'), ω' becomes 4ω. A faster rotation means a shorter day: T' = T/4 = 24/4 = 6 hours.

Tangential acceleration acts along the tangent to the path, thus changing the object's speed.

Moment of inertia is an intrinsic property of an object's mass distribution, independent of its motion.

The formula for orbital velocity is v_o = √(GM/R) and for escape velocity is v_e = √(2GM/R). Therefore, v_e = √2 v_o.

At the top, both gravity and the normal force (if needed) act downwards, providing the centripetal force to keep the water moving in a circle.

Because the hoop has a larger moment of inertia, it will have a smaller translational velocity compared to the disk.

Total linear acceleration is the vector sum of the tangential component (changing speed) and the centripetal component (changing direction).

Rotational equilibrium means the net torque is zero, so there is no change in angular velocity (i.e., zero angular acceleration).

KE_rot = (1/2)Iω² = (1/5)Mv². KE_total = (7/10)Mv². The ratio is (1/5)/(7/10) = 2/7.

This is the rotational equivalent of Newton's second law: τ = dL/dt.

At the top, the tangential velocity (v) and the translational velocity (v) are in the same direction, so the resultant velocity is v + v = 2v.

For rotational equilibrium, there must be no net turning effect, meaning the sum of all torques is zero.

A spinning object has angular momentum. To change its direction (i.e., to make it fall), a torque is required. Rapid spinning makes it resistant to such torques.

A geostationary orbit allows a satellite to remain over the same point on Earth's surface, which is ideal for continuous communication.

At the equator, part of the gravitational force provides the centripetal force for rotation. The apparent weight is thus reduced: W_app = mg - mω²r.

Since one revolution is 2π radians, three revolutions is 3 × 2π = 6π radians.

Artificial gravity is often measured in multiples of Earth's gravitational acceleration. '2g' means an acceleration of 2 × 9.8 m/s².

The negative potential energy is greater in magnitude than the positive kinetic energy, signifying a bound system.

Since τ = ΔL/Δt, then Δt = ΔL/τ = 10 J·s / 2 N·m = 5 s.

By the perpendicular axis theorem, I_z = I_x + I_y. For a disk, I_x = I_y and I_z = (1/2)MR². So, 2I_x = (1/2)MR², which gives I_x = (1/4)MR².

Using the right-hand rule, if you curl your fingers counter-clockwise, your thumb points out of the page.

This is the rotational work formula, analogous to linear work W = Fd.

The period (T) is 24 hours = 86400 s. Frequency is f = 1/T, so f = 1/86400 Hz.

To maintain a stable orbit at a greater distance, less orbital speed is required to balance the weaker gravitational pull.

In general relativity, massive objects warp spacetime, and this curvature is what we perceive as gravity.

First, ωᵢ = 1200 × (2π/60) = 40π rad/s. Then, α = (ω_f - ωᵢ)/t = (0 - 40π)/10 = -4π rad/s². The magnitude is 4π rad/s².

Using τ = Iα, we get α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².

Angular momentum is L = r × p. If the particle moves along a line passing through the reference point, r is parallel to p, so their cross product L is zero and conserved.

Planck's constant has units of Joule-seconds (J·s), which gives dimensions of [ML²T⁻²]·[T] = [ML²T⁻¹], the same as angular momentum.

From F = G(m₁m₂/r²), we rearrange to get G = Fr²/(m₁m₂). The dimensions are [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²].

This is the definition of the cross product τ = r × F. The direction is given by the right-hand rule.

This arrangement produces a pure torque (turning effect) without any net translational force.

As the radius increases, the moment of inertia (I ∝ R²) increases. Since angular momentum (L = Iω) is conserved, the angular velocity (ω) must decrease.

Using θ = ωᵢt + (1/2)αt², with ωᵢ = 0, we get θ = (1/2)(2)(4)² = 16 rad.

Just as linear impulse equals the change in linear momentum, angular impulse (τΔt) equals the change in angular momentum (ΔL).

At the top, the minimum centripetal force is provided by gravity alone (tension T=0). So, mg = mv²/R, which gives v = √(gR).

This is the rotational equivalent of the linear power formula P = Fv.

This condition ensures that the point of contact with the surface is instantaneously at rest.

The original acceleration is a = rω². The new acceleration is a' = (r/2)(2ω)² = (r/2)(4ω²) = 2(rω²) = 2a.

We need g' = g/2. So, GM/(R+h)² = GM/(2R²). This gives (R+h)² = 2R², so R+h = √2 R, and h = (√2-1)R.

Due to its large moment of inertia, a flywheel resists changes in angular speed, storing and releasing energy to maintain a nearly constant rotation.

According to Kepler's Third Law as derived by Newton, T² = (4π²/GM_sun)r³. The period depends on the mass of the central body.

The gear ratio is 40:10 or 4:1. For every 4 revolutions of the smaller gear, the larger gear makes 1 revolution.

The term 'g-force' refers to the total acceleration felt by the pilot, which creates a sensation of increased apparent weight.

For uniform circular motion, the magnitude (Iω) and direction (along the axis) of the angular momentum vector are both constant.

Static friction acts horizontally, towards the center of the turn, preventing the car from skidding outwards.

In free fall, an object experiences no support force, leading to the sensation of weightlessness, even though gravity is still acting on it.

Moment of inertia is defined as I = Σmᵢrᵢ², so its units are mass (kg) times distance squared (m²).

In free fall, the effective gravity (g_eff) is zero. Since T = 2π√(L/g_eff), as g_eff → 0, T → ∞. The pendulum will not oscillate.

This is the fundamental condition for the conservation of angular momentum.

A hollow sphere, a ring, or an L-shaped object can have a center of mass located in empty space.

She decreases her moment of inertia (I). To conserve angular momentum (L=Iω), her angular velocity (ω) must increase.

The Earth is slightly flattened, making the poles closer to the center. This, plus the lack of rotational effect, makes 'g' maximum at the poles.

Centripetal force is the real net force causing circular motion. Centrifugal force is an apparent outward force in a non-inertial (rotating) frame.

Escape velocity is v_e = √(2GM/R). The Moon's smaller mass (M) and radius (R) both contribute to a lower escape velocity.

33 rev/min × (2π rad/rev) × (1 min/60 s) = 66π/60 ≈ 3.46 rad/s.

This is the rotational analogue of the linear work-energy theorem. Here, W_torque = ΔKE_rotational.

By conservation of angular momentum (L=mvr), when the radius 'r' is smallest, the velocity 'v' must be largest.

The equation is ω_f = ωᵢ + αt. With a constant negative α, this is the equation of a straight line (y = c + mx) with a negative slope.