Circular Motion: Practice Questions (Physics 1 2.9)

Answer: (C). The frequency is 4.0 Hz, so the tip travels 4.0 circumferences per second: v = 2πr f = 2π × 0.25 m × 4.0 Hz = 6.28 m/s. Then a_c = v² / r = (6.28)² ÷ 0.25 = 158 m/s² ≈ 160 m/s².

• (A) uses v = r f = 1.0 m/s, forgetting that one revolution is a distance of 2πr, not r.
• (B) is v / r = 25. It forgets to square the speed (and has the wrong units: 1/s, not m/s²).
• (D) multiplies by r instead of dividing: v² × r = 9.9. A quick unit check ((m/s)² × m = m³/s²) rules it out.

Answer: (C). For circular orbits around the same central mass, T² ∝ r³. So (T_Y / T_X)² = 4³ = 64, and T_Y / T_X = √64 = 8.

• (A) uses T ∝ √r, a square root in the wrong place.
• (B) assumes T ∝ r. That needs equal speeds, but the outer moon moves more slowly.
• (D) uses T ∝ r², squaring instead of taking T² ∝ r³.

The masses of the moons are not needed: they cancel when gravity supplies the centripetal acceleration.

Answer: (C). The car has two perpendicular parts of acceleration. The centripetal part points to the centre, which is due south from the northern point. The tangential part points forward (west) because the car is speeding up. Their vector sum points between south and west.

• (A) would be correct only at constant speed, when the tangential part is zero.
• (B) ignores the change in direction. A car moving on a curve always has a centripetal part.
• (D) points partly outward. Neither part of the acceleration points away from the centre.

At the top, the centre of the circle is below the car. Take “towards the centre” (downward) as positive. Weight mg points down (inward); the normal force N points up (outward).

(a) Newton’s second law inward: mg − N = m v² / r. The car loses contact when N falls to zero, so the greatest speed has mg = m v² / r: v_max = √(g r) = √(9.8 × 30) = √294 = 17 m/s.

(b) N = m(g − v² / r) = 1200 kg × (9.8 − 144 ÷ 30) m/s² = 1200 × (9.8 − 4.8) = 1200 × 5.0 = 6.0 × 10³ N, upward.

Check. The normal force (6000 N) is less than the weight (11 760 N). It must be, because the net force has to point down, towards the centre.

Suggested mark points (4): 1 for a correct inward equation with mg and N in opposite directions; 1 for setting N = 0 at the greatest speed; 1 for 17 m/s; 1 for 6.0 × 10³ N with direction.

Common error: writing N − mg = m v²/r, as at the bottom of a dip. That gives N = 17 520 N, more than the weight, but on a crest the net force must point down.

(a) With no friction: vertical N cos θ = mg; inward N sin θ = m v² / r. Dividing gives tan θ = v² / (r g), so v = √(r g tan θ) = √(25 × 9.8 × tan 40°) = √(25 × 9.8 × 0.839) = √206 = 14.3 m/s ≈ 14 m/s.

(b) Friction acts up the slope. At 10 m/s the cyclist needs only m v²/r = m × 4.0 m/s² inward, but the horizontal part of the normal force on this steep bank would provide more than that (it gives m × 8.2 m/s² when friction is zero). Without friction the cyclist would slide down and in towards the centre, so static friction acts up the slope to prevent it.

(c) From v = √(r g tan θ), with θ and g unchanged, v ∝ √r. Doubling r multiplies v by √2 ≈ 1.4 (about 20 m/s).

Suggested mark points (5): 1 for both component equations (N cos θ = mg and N sin θ = m v²/r); 1 for tan θ = v²/(rg) or the equivalent, leading to about 14 m/s; 1 for friction up the slope; 1 for the reason (needed inward force is less than N sin θ would give, so it tends to slide down); 1 for √2.

With friction on a bank, the course asks only for descriptions like part (b).

(a) Reaction time adds roughly the same error to every timing. Spreading that error over 10 revolutions makes the error in each period about ten times smaller.

(b) Plot F (vertical axis, N) against v² (horizontal axis, m²/s²). If F = (m / r) v², the points lie on a straight line through the origin with slope m / r.

(c) For each run: T = time ÷ 10, then v = 2πr / T.

Using points far apart on the best-fit line (here close to runs 1 and 5): slope = (2.48 − 0.63) N ÷ (15.42 − 3.86) m²/s² = 1.85 ÷ 11.56 = 0.160 kg/m. A computer best fit gives 0.160 kg/m with an intercept of almost zero.

(d) slope = m / r, so m = slope × r = 0.160 kg/m × 0.50 m = 0.080 kg. This matches the balance reading, and the points lie close to a straight line through the origin, so the data support F = m v² / r.

Alternative method. Plotting F against 1/T² also gives a straight line, with slope 4π² m r. Here that is 4π² × 0.080 × 0.50 = 1.58 N·s². It earns full credit when the slope is linked to m correctly. Plotting F against v gives a curve and earns no credit for (b).

(a) Gravity is the only force, and it points to the centre, so: G M m / r² = m v² / r. Cancel m and one r: v² = G M / r. For constant speed, v = 2πr / T, so 4π² r² / T² = G M / r, giving T² = 4π² r³ / (G M), or T = 2π √(r³ / (G M)).

(b) The claim is incorrect. The heavier moon does feel a gravitational force ten times larger, but it also has ten times the mass, so its acceleration (F ÷ m) is the same. In (a) the moon’s mass m cancels, and T depends only on r and M. Same radius and same planet means the same period.

(c) M = 4π² r³ / (G T²) = 39.5 × (2.0 × 10⁸)³ ÷ [(6.67 × 10⁻¹¹) × (5.0 × 10⁵)²] = (3.16 × 10²⁶) ÷ 16.7 = 1.9 × 10²⁵ kg.

An answer to (b) that says only “mass doesn’t matter” without referring to the cancellation or to equal acceleration earns no credit for the third point.