Circular motion – A-level Physics exam questions with mark scheme

1Core3 marks

The blades of a wind turbine have a length of 42 m. The blades rotate at a constant rate of 15 revolutions per minute.

(a) Calculate the angular speed of the blades. [1]

(b) Calculate the linear speed of a blade tip. [1]

(a) $\omega = 2\pi f = 2\pi \times \frac{15}{60}$ = 1.6 rad s$^{-1}$ (1.57) allow $\pi/2$(1)
(b) $v = \omega r = 1.57 \times 42$ = 66 m s$^{-1}$ (ecf from (a))(1)
(c) $a = \omega^2 r$ or $v^2/r$ = 104 m s$^{-2}$ (103.6; allow 100 to 2 sf) (ecf)(1)
Answer: (a) 1.57 rad s$^{-1}$ (b) 66 m s$^{-1}$ (c) 104 m s$^{-2}$
Examiner tip: Convert revolutions per minute to revolutions per second before multiplying by 2π — leaving it per minute is the most common slip.

2Standard4 marks

A car travels over the top of a hump-backed bridge. The top of the bridge is an arc of a vertical circle of radius 25 m.

(a) Explain why the normal contact force on the car from the road is less than the weight of the car when the car is at the top of the bridge. [2]

(b) Calculate the maximum speed at which the car can travel over the top of the bridge without losing contact with the road. [2]

(a) a resultant (centripetal) force is needed towards the centre of the circle, i.e. downwards(1)
(a) resultant force = weight − normal force $= mv^2/r$, so normal force $= mg - mv^2/r$ which is less than $mg$(1)
(b) contact lost when normal force = 0, so $mg = mv^2/r$ or $v = \sqrt{gr}$(1)
(b) $v = \sqrt{9.81 \times 25}$ = 16 m s$^{-1}$ (15.7)(1)
Answer: (b) 15.7 m s$^{-1}$
Examiner tip: Say which force is bigger and why: the weight must exceed the reaction to leave a downward resultant towards the centre.

3Hard5 marks

A small ball of mass 0.20 kg is attached to a string of length 0.50 m. The ball moves at constant speed in a horizontal circle with the string at 30° to the vertical, as shown.

(a) Show that the tension in the string is about 2.3 N. [2]

(b) Calculate the speed of the ball. [3]

(a) vertical forces balance: $T\cos 30^\circ = mg$(1)
(a) $T = \frac{0.20 \times 9.81}{\cos 30^\circ}$ = 2.27 N (at least 3 sf seen for a 'show that')(1)
(b) radius of circle $r = 0.50 \sin 30^\circ = 0.25$ m(1)
(b) horizontal component of tension provides centripetal force: $T \sin 30^\circ = \frac{mv^2}{r}$ (or $v^2 = gr\tan 30^\circ$)(1)
(b) $v$ = 1.2 m s$^{-1}$ (1.19) (allow use of 2.3 N giving 1.20)(1)
Answer: (a) 2.27 N (b) 1.19 m s$^{-1}$
Examiner tip: The radius of the circle is not the string length — resolve to get r = L sin θ.

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4Hard7 marks

A ball of mass 0.15 kg is attached to a light string and whirled in a vertical circle of radius 0.80 m. At the top of the circle the speed of the ball is 3.5 m s$^{-1}$. Assume no energy is lost to air resistance.

(a) Calculate the tension in the string when the ball is at the top of the circle. [2]

(b) Calculate the speed of the ball at the bottom of the circle. [2]

(d) Explain why the string is most likely to break when the ball is at the bottom of the circle. [1]

(a) at top: $T + mg = \frac{mv^2}{r}$(1)
(a) $T = 0.15\left(\frac{3.5^2}{0.80} - 9.81\right)$ = 0.83 N (0.825)(1)
(b) loss in GPE = gain in KE: $v_b^2 = v_t^2 + 2g(2r) = 3.5^2 + 2 \times 9.81 \times 1.60$(1)
(b) $v_b$ = 6.6 m s$^{-1}$ (6.61)(1)
(c) at bottom: $T - mg = \frac{mv^2}{r}$(1)
(c) $T = 0.15\left(9.81 + \frac{6.61^2}{0.80}\right)$ = 9.7 N (9.65) (allow 9.6 N from $v$ = 6.6 m s$^{-1}$; ecf from (b))(1)
(d) at the bottom the tension must both support the weight and provide the centripetal force (and speed is greatest), so tension is greatest there(1)
Answer: (a) 0.83 N (b) 6.6 m s$^{-1}$ (c) 9.7 N
Examiner tip: The height change from top to bottom is the diameter (2r), not the radius.

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