Rotational And Circular Motion MCQs | Physics MCQs

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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.

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.

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

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.

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

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

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.

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.

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

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.

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

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

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

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

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

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

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

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.

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

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

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

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.

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.

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

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

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

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 rotational equivalent of the linear power formula P = Fv.

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

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

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

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.

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

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.

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

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

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.

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

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

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

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

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

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

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

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

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

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.

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

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.

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