Year 12 Physics Module 5: Advanced Mechanics IQ2: Circular Motion Lessons 6–12 45 min

Checkpoint 2: Circular Motion

Test the prescribed core from Lessons 6–8 and 10–12: torque, uniform circular motion, banked curves, gravitation, orbital mechanics and energy in circular orbits. Lesson 9 vertical circles are optional extension and are not scored here.

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Topics Covered

Lessons 6–7

Torque & Uniform Circular Motion

$a_c = v^2/r = \omega^2 r$ and $F_c = mv^2/r = m\omega^2 r$
Angular velocity $\omega = 2\pi/T = 2\pi f$
Identifying the source of centripetal force

Lesson 8 · L09 is extension

Horizontal Circular Motion

Conical pendulum: $\tan\theta = v^2/rg$
Banked curves: $v_{\text{design}} = \sqrt{rg\tan\theta}$
Friction direction from impending relative motion
Radial force does zero work in ideal uniform motion

Lessons 10–12

Newton's Law, Orbital Mechanics & Energy in Orbits

Orbital velocity: $v = \sqrt{GM/r}$
Kepler's Third Law: $T^2 \propto r^3$
Geostationary orbits ($T = 24$ h, $h \approx 35\,900$ km)
Energy: $KE = \frac{1}{2}GMm/r$, $U = -GMm/r$, $E = -\frac{1}{2}GMm/r$
Escape velocity: $v_e = \sqrt{2GM/r}$

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Formula Reference

$\tau = rF_{\perp} = rF\sin\theta$

$\tau$ = torque (N m), $r$ = lever arm (m), $F$ = force (N), $\theta$ = angle between $r$ and $F$

$a_c = \dfrac{v^2}{r} = \omega^2 r$

$a_c$ = centripetal acceleration (m s^-2)

$v$ = linear speed (m s^-1), $r$ = radius (m), $\omega$ = angular velocity (rad s^-1)

$F_c = \dfrac{mv^2}{r} = m\omega^2 r$

$F_c$ = centripetal force (N), $m$ = mass (kg)

$v = \omega r$, $\quad \omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{2\pi}{T} = 2\pi f$

$T$ = period (s), $f$ = frequency (Hz)

$\tan\theta = \dfrac{v^2}{rg}$

Conical pendulum & banked curves (no friction). $\theta$ = angle to vertical / banking angle

$W_{\text{radial}} = F_c d\cos 90^\circ = 0$

In ideal uniform circular motion, radial force is perpendicular to instantaneous displacement, so it changes direction but not speed.

$F_c\propto m$, $\quad F_c\propto v^2$, $\quad F_c\propto 1/r$

Hold the other variables constant when interpreting each relationship.

$v = \sqrt{\dfrac{GM}{r}}$

Orbital speed. $G$ = $6.674 \times 10^{-11}$ N m² kg^-2, $M$ = central mass (kg)

Remember: $r = R_{\text{planet}} + h$ (centre-to-centre distance)

$\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM}$

Kepler's Third Law for circular orbits

$KE = \dfrac{GMm}{2r}$, $\quad U = -\dfrac{GMm}{r}$, $\quad E = -\dfrac{GMm}{2r}$

Kinetic, gravitational potential, and total mechanical energy in orbit

$v_e = \sqrt{\dfrac{2GM}{r}}$

Escape velocity from a body's surface (or any distance $r$ from centre)

$g = \dfrac{GM}{r^2}$

Gravitational field strength at distance $r$ from centre. Use for orbital altitudes.

Multiple Choice Questions

Multiple Choice, 17 Questions

Select the best answer for each question. Feedback appears after you choose.

Q1. A car moves around a circular track at constant speed. Which statement is correct?

The car has zero acceleration because its speed is constant.

The car is accelerating because the direction of its velocity is continuously changing.

The car is accelerating because its speed is increasing.

The car experiences no net force because it moves at constant speed.

Correct! In uniform circular motion, although speed is constant, the direction of velocity continuously changes. This means there is a change in velocity, hence acceleration (centripetal acceleration) directed toward the centre.

Incorrect. Remember that velocity is a vector (speed + direction). Even at constant speed, a change in direction means a change in velocity, which is acceleration. Also, circular motion requires a net inward force.

Q2. A particle moves in a circle of radius 2.0 m with angular velocity $4.0\text{ rad s}^{-1}$. What is its centripetal acceleration?

$2.0\text{ m s}^{-2}$

$8.0\text{ m s}^{-2}$

$32\text{ m s}^{-2}$

$16\text{ m s}^{-2}$

Correct! Using $a_c = \omega^2 r = (4.0)^2 \times 2.0 = 16 \times 2.0 = 32\text{ m s}^{-2}$.

Incorrect. Use $a_c = \omega^2 r$ (not $\omega r$). With $\omega = 4.0\text{ rad s}^{-1}$ and $r = 2.0\text{ m}$: $a_c = (4.0)^2 \times 2.0 = 32\text{ m s}^{-2}$.

Q3. A 0.50 kg mass is attached to a string and whirled in a horizontal circle of radius 0.80 m at $3.0\text{ m s}^{-1}$. What is the tension in the string?

$1.9\text{ N}$

$3.4\text{ N}$

$5.6\text{ N}$

$7.5\text{ N}$

Correct! Tension provides the centripetal force: $T = F_c = mv^2/r = (0.50)(3.0)^2/(0.80) = 4.5/0.80 = 5.625 \approx 5.6\text{ N}$.

Incorrect. The tension equals the centripetal force required: $F_c = mv^2/r = (0.50)(9.0)/(0.80) = 5.6\text{ N}$. Check your substitution and division.

Q4. A satellite completes one orbit every 90 minutes. What is its angular velocity in $\text{rad s}^{-1}$?

$4.7 \times 10^{-4}\text{ rad s}^{-1}$

$1.2 \times 10^{-3}\text{ rad s}^{-1}$

$2.1 \times 10^{-2}\text{ rad s}^{-1}$

$6.9 \times 10^{-4}\text{ rad s}^{-1}$

Correct! $\omega = 2\pi/T$. With $T = 90 \times 60 = 5400\text{ s}$: $\omega = 2\pi/5400 = 1.16 \times 10^{-3}\text{ rad s}^{-1}$.

Incorrect. Convert 90 minutes to seconds first: $T = 90 \times 60 = 5400\text{ s}$. Then $\omega = 2\pi/T = 2\pi/5400 \approx 1.16 \times 10^{-3}\text{ rad s}^{-1}$.

Q5. A mechanic applies a 60 N force to a wrench of length 0.30 m at an angle of 30° to the wrench handle. What is the torque applied to the nut?

$9.0\text{ N m}$

$18\text{ N m}$

$15.6\text{ N m}$

$4.5\text{ N m}$

Correct! $\tau = rF\sin\theta = 0.30 \times 60 \times \sin 30^{\circ} = 0.30 \times 60 \times 0.5 = \mathbf{9.0\text{ N m}}$.

Incorrect. Use $\tau = rF\sin\theta$. Here $r = 0.30\text{ m}$, $F = 60\text{ N}$, and $\theta = 30^{\circ}$ (the angle between the lever arm and the force). So $\tau = 0.30 \times 60 \times \sin 30^{\circ} = 0.30 \times 60 \times 0.5 = 9.0\text{ N m}$. Note that only the perpendicular component of force contributes to torque.

Q6. A uniform seesaw is balanced horizontally with a 30 kg child sitting 2.0 m from the pivot on one side. Where must a 40 kg child sit on the other side to maintain balance?

$1.0\text{ m}$

$1.3\text{ m}$

$1.5\text{ m}$

$2.7\text{ m}$

Correct! For rotational equilibrium, clockwise torque = anticlockwise torque. Both children sit perpendicular to the seesaw ($\theta = 90^{\circ}$, so $\sin\theta = 1$). $m_1 g r_1 = m_2 g r_2$, giving $r_2 = (30 \times 2.0) / 40 = \mathbf{1.5\text{ m}}$.

Incorrect. For balance, the torques must be equal: $\tau_1 = \tau_2$. Since both children sit perpendicular to the seesaw, $\tau = rF = rmg$. So $r_1 m_1 g = r_2 m_2 g$, which simplifies to $r_2 = r_1 m_1 / m_2 = (2.0 \times 30) / 40 = 1.5\text{ m}$.

Q7. In a conical pendulum, what provides the centripetal force?

The full tension in the string

The horizontal component of tension

The vertical component of tension

The weight of the bob

Correct! The horizontal component of tension, $T\sin\theta$, provides the centripetal force that keeps the bob moving in a circle. The vertical component $T\cos\theta$ balances the weight $mg$.

Incorrect. Only the horizontal component of tension points toward the centre of the circular path. The vertical component balances weight. Resolve $\vec{T}$ into components to analyse.

Q8. A conical pendulum makes an angle $\theta$ with the vertical. Which relationship is correct?

$\tan\theta = v^2/(rg)$

$\sin\theta = v^2/(rg)$

$\cos\theta = v^2/(rg)$

$\tan\theta = rg/v^2$

Correct! From resolving forces: vertically $T\cos\theta = mg$, horizontally $T\sin\theta = mv^2/r$. Dividing: $\tan\theta = v^2/(rg)$.

Incorrect. Resolve forces: vertical gives $T\cos\theta = mg$, horizontal gives $T\sin\theta = mv^2/r$. Dividing the second by the first: $\tan\theta = v^2/(rg)$. Note the correct position of $v^2$ and $rg$.

Q9. A curve of radius 50 m is banked at $15^{\circ}$. What is the design speed at which no friction is needed? ($g = 9.8\text{ m s}^{-2}$)

$8.2\text{ m s}^{-1}$

$11.5\text{ m s}^{-1}$

$16.3\text{ m s}^{-1}$

$22.1\text{ m s}^{-1}$

Correct! Using $v = \sqrt{rg\tan\theta} = \sqrt{50 \times 9.8 \times \tan 15^{\circ}} = \sqrt{50 \times 9.8 \times 0.268} = \sqrt{131.3} \approx 11.5\text{ m s}^{-1}$.

Incorrect. Use $v_{\text{design}} = \sqrt{rg\tan\theta}$. Remember to use $\tan$, not $\sin$ or $\cos$. With $r = 50\text{ m}$, $g = 9.8\text{ m s}^{-2}$, $\theta = 15^{\circ}$: $v = \sqrt{50 \times 9.8 \times 0.268} \approx 11.5\text{ m s}^{-1}$.

Q10. A car travels faster than the design speed on a banked curve. What happens?

The car slides down the bank

Friction acts down the slope to provide additional centripetal force

The normal force becomes zero

The car automatically slows to the design speed

Correct! Above the design speed, the horizontal component of the normal force alone is insufficient. Friction acting down the slope adds an extra horizontal component toward the centre, increasing the centripetal force.

Incorrect. At speeds above the design speed, the car tends to slide up the bank. Friction therefore acts down the slope, and its horizontal component adds to the normal force's horizontal component to provide more centripetal force.

Q11. Why does the radial net force do zero work on an object in ideal uniform circular motion?

The radial net force is zero

The object has no displacement

The radial net force is perpendicular to the instantaneous displacement

The kinetic energy is zero

Correct! Instantaneous displacement is tangential while the radial resultant points inward, so $W=Fd\cos90^\circ=0$. It redirects velocity without changing its magnitude.

Incorrect. A nonzero radial force acts while the object moves, but force and instantaneous displacement are perpendicular. Therefore $W=Fd\cos90^\circ=0$.

Q12. An object moves in a circle at twice its original speed while mass and radius remain constant. The required radial net force becomes:

Half as large

Twice as large

Four times as large

Unchanged

Correct! Since $F_c=mv^2/r$, replacing $v$ by $2v$ gives $F_c'=m(2v)^2/r=4F_c$.

Incorrect. Speed is squared in $F_c=mv^2/r$. Doubling speed therefore multiplies the required radial net force by four.

Q13. A satellite orbits Earth at altitude $h$. Which expression gives the correct orbital speed? ($R_E$ = Earth's radius)

$v = \sqrt{GM_E/h}$

$v = \sqrt{GM_E/(R_E + h)}$

$v = \sqrt{GM_E/R_E}$

$v = \sqrt{gh}$

Correct! The orbital speed depends on the centre-to-centre distance $r = R_E + h$, not just the altitude. From $GM_E m/r^2 = mv^2/r$, we get $v = \sqrt{GM_E/r} = \sqrt{GM_E/(R_E + h)}$.

Incorrect. Remember that $r$ in the orbital velocity formula is the centre-to-centre distance: $r = R_E + h$. Also, $g = 9.8\text{ m s}^{-2}$ is not valid at altitude; you must use $g = GM_E/r^2$ at that location.

Q14. Two satellites orbit the same planet. Satellite A has orbital radius $r$ and period $T$. Satellite B has orbital radius $4r$. What is the period of satellite B?

$2T$

$4T$

$8T$

$16T$

Correct! By Kepler's Third Law: $T^2 \propto r^3$, so $(T_B/T_A)^2 = (r_B/r_A)^3 = (4)^3 = 64$. Therefore $T_B/T_A = 8$, giving $T_B = 8T$.

Incorrect. Kepler's Third Law states $T^2 \propto r^3$. So $(T_B/T)^2 = (4r/r)^3 = 4^3 = 64$, and $T_B/T = \sqrt{64} = 8$. The period increases by factor 8, not 4 or 16.

Q15. Which of the following correctly describes the total mechanical energy of a satellite in a stable circular orbit?

Positive, because the satellite is moving

Zero, because kinetic and potential energy cancel

Negative, equal to $-\dfrac{GMm}{2r}$

Negative, equal to $-\dfrac{GMm}{r}$

Correct! $E = KE + U = \frac{1}{2}{G}{M}{m}/r + (-{G}{M}{m}/r) = -\frac{1}{2}{G}{M}{m}/r$. The total energy is negative (indicating a bound orbit) and has magnitude equal to the kinetic energy.

Incorrect. $U = -GMm/r$ and $KE = +\frac{1}{2}GMm/r$. So $E = KE + U = -\frac{1}{2}GMm/r$. The factor of $\frac{1}{2}$ is essential. Negative total energy means the satellite is gravitationally bound.

Q16. A geostationary satellite must satisfy which condition?

It orbits at any altitude with period 24 hours

It orbits above the equator with period 24 hours in the same direction as Earth's rotation

It orbits over the poles with period 12 hours

It orbits at altitude $35\,900$ km with any period

Correct! A geostationary satellite must: (1) orbit above the equator, (2) have period $T = 24$ hours matching Earth's rotation, and (3) orbit in the same direction (prograde). These conditions fix the altitude at approximately $35\,900$ km.

Incorrect. All three conditions are necessary: equatorial orbit, $T = 24$ h, and prograde direction. The altitude is a consequence of the period requirement, not an independent choice. Polar orbits cannot be geostationary.

Q17. Optional preview of Lesson 16, not part of the core score. The escape speed from a planet depends on which factors?

The planet's mass and the projectile's mass

The planet's mass and the distance from the planet's centre

The projectile's mass and launch speed only

The planet's radius only

Correct! From $v_e = \sqrt{2GM/r}$, escape velocity depends on the planet's mass $M$ and the distance $r$ from its centre. The projectile's mass cancels out in the derivation. For surface escape, $r = R_{\text{planet}}$.

Incorrect. $v_e = \sqrt{2GM/r}$ shows that escape velocity depends on the central body's mass $M$ and the radial distance $r$, but not on the projectile's mass (it cancels in the energy equation). It depends on radius only if launching from the surface.

Short Answer Questions

Short Answer, 5 Questions

Show all working. Marks are awarded for correct method, substitution, and final answer with units.

Q1. A 1200 kg car travels around a 35 m radius curve at 18 m/s. Calculate the centripetal acceleration and the friction force required. Explain what happens if the road is icy. 3 marks