2.9 Circular Motion — Worked Solutions
Work through Topic 2.9
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A puck moves uniformly in a circle. Which quantity is zero?
- Centripetal acceleration
- Tangential acceleration
- Net acceleration
Step 1. Uniform circular motion keeps speed constant but turns the velocity direction.
Why this step? Uniform refers to speed around the circular path, not to a fixed velocity vector.
Step 2. Tangential acceleration is zero because the speed does not change.
Why this step? This component measures the rate of change of speed along the tangent, which is zero under the stated uniform condition.
Step 3. Centripetal and net acceleration remain nonzero, with magnitude v²/r for a nonzero speed.
Why this step? The direction change still requires inward acceleration, so neither radial nor total acceleration is the zero choice.
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Question 2
At the minimum contact speed at the top of an inside vertical loop, what supplies the inward force?
- Gravity alone
- Gravity plus an extra centripetal force
- A nonzero normal force only
Step 1. At the top, inward is downward, so the force equation is mg + N = mv²/r.
Why this step? Both gravity and an inside-track normal force point toward the center there; “centripetal” is not an extra interaction.
Step 2. At the minimum contact speed, set N = 0 and solve mg = mv²/r, giving v_min = √(gr).
Why this step? A normal force cannot pull the object onto the inside track; the threshold is the lowest speed needing no such impossible pull.
Step 3. Gravity alone supplies the inward force at that threshold.
Why this step? Below the threshold the required normal would be negative, so the circular contact condition would fail instead of producing a new force.
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Question 3
Circular orbital radius doubles around the same central body. How does period change?
- It doubles
- It grows by 2√2
- It halves
Step 1. For the same central mass, P² = [4π²/(GM)]r³.
Why this step? The gravitational parameter GM is fixed across the two circular orbits, so only the orbital radius changes the period ratio.
Step 2. Double r: (P_new/P_old)² = (2r/r)³ = 8.
Why this step? The cubed radius factor acts on the squared period, so the factor is not simply two.
Step 3. Take the positive time ratio: P_new/P_old = √8 = 2√2.
Why this step? Period is a positive elapsed time; taking the square root translates the squared-period relation into the requested period change.
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Question 4
A 0.50 kg mass moves uniformly at 4 m/s in a circle of radius 2 m. Find inward acceleration, inward net force, and period. In a separate changing-speed case, its instantaneous speed is 4 m/s at the same radius and its tangential acceleration is 6 m/s². Find the total acceleration magnitude at that instant.
Step 1. Find inward acceleration a_c = v²/r = 4²/2 = 8 m/s².
Why this step? Circular curvature fixes the radial acceleration at the stated instantaneous speed and radius, independent of which real force supplies it.
Step 2. The inward net-force component is ma_c = 0.50·8 = 4 N.
Why this step? Newton’s second law uses the given mass to convert radial acceleration into the radial resultant, not an additional force arrow.
Step 3. If the first case maintains speed 4 m/s throughout a revolution, P = 2πr/v = 2π·2/4 = π s.
Why this step? A complete lap has length 2πr; dividing by one speed gives its duration only if that speed stays constant.
Step 4. With tangential acceleration 6 m/s² at this instant, total magnitude is √(8² + 6²) = 10 m/s².
Why this step? Radial and tangential components are perpendicular, so Pythagoras rather than scalar addition combines them.
Step 5. If the speed changes over a lap, the instantaneous 4 m/s alone cannot determine that lap’s period.
Why this step? The added tangential acceleration changes future speed; a full time history would be needed for a nonuniform revolution.
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Question 5
Derive tan θ = v²/(rg) for a frictionless banked curve. Explain why the normal force must exceed mg whenever θ is nonzero.
Step 1. The tilted normal force has vertical component N cos θ and inward horizontal component N sin θ.
Why this step? For bank angle θ, the normal tilts by θ from vertical; the frictionless assumption leaves no other contact force to supply horizontal acceleration.
Step 2. Write N cos θ = mg and N sin θ = mv²/r.
Why this step? The car has no vertical acceleration but does have inward horizontal acceleration while rounding the level-radius curve.
Step 3. Divide the second equation by the first: tan θ = v²/(rg).
Why this step? The unknown normal force and mass cancel in the component ratio, directly linking design angle, speed, and radius.
Step 4. For 0 < θ < 90°, N = mg/cos θ > mg.
Why this step? Only the vertical component balances weight; since cos θ is less than one, the full tilted force must be larger than that component.
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Question 6
Derive the circular-orbit period equation by eliminating speed. Sketch P² versus r³ for satellites around one planet and identify the slope, including its units.
Step 1. Set GMm/r² = mv²/r, then cancel m to obtain v² = GM/r.
Why this step? Gravity is the only inward force in the circular-orbit model; satellite mass occurs on both sides and does not set orbital speed.
Step 2. For uniform circular orbit, P = 2πr/v. Square it: P² = 4π²r²/v².
Why this step? One revolution covers the circumference at constant speed; squaring prepares the expression for substitution of v².
Step 3. Substitute v² = GM/r to get P² = 4π²r³/(GM).
Why this step? Dividing by GM/r introduces one extra factor of r, giving the cubic-radius rather than quadratic-radius dependence.
Step 4. Plot P² against r³: the line has zero intercept and slope 4π²/(GM), with units s²/m³.
Why this step? The same planet keeps GM fixed, so the coefficient is constant; the vertical-to-horizontal axis-unit ratio checks its dimensions.
Step 5. A larger central mass gives a smaller slope.
Why this step? GM is in the denominator: at the same orbital radius, stronger gravity allows a faster orbit and therefore a shorter period.