Motion in a circle — WME03 IAL Further Mathematics practice questions · Study-IAL

Motion in a circle

WME03Topic 3.43 questions

Uniform motion in a horizontal circle and motion in a vertical circle.

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Original practice questions written in the style of the exam.

Question 15 marks
A particle of mass $0.3\,\text{kg}$ is attached to one end of a light inextensible string of length $1.2\,\text{m}$ . The other end is fixed. The particle moves in a horizontal circle with the string making a constant angle of $30^\circ$ with the vertical.

Find (a) the tension in the string and (b) the angular speed of the particle.
Mark scheme
- M1 Resolves vertically: $T\cos 30^\circ = 0.3 \times 9.8$ (1)
- A1 $T = 3.39\,\text{N}$ (3 s.f.) (1)
Question 26 marks
A particle of mass $0.5\,\text{kg}$ is attached to one end of a light inextensible string of length $0.8\,\text{m}$ , the other end of which is attached to a fixed point $O$ . The particle is at its lowest point, moving horizontally with speed $7\,\text{m s}^{-1}$ , and then moves in a vertical circle about .

B1 Radius

r = 1.2\sin 30^\circ = 0.6\,\text{m}

(1)

M1 Horizontally:

T\sin 30^\circ = 0.3 \times 0.6\,\omega^2

(1)

\omega = 3.07\,\text{rad s}^{-1}

(3 s.f.) (1)

−1

O

(a) Find the tension in the string when the string is horizontal.

(b) Find the tension in the string when the particle is at the top of the circle, and hence explain why the particle completes the circle.

Mark scheme

M1 Energy: $v^2 = 49 - 2 \times 9.8 \times 0.8 = 33.32$ when the string is horizontal (1)
A1 $T = \dfrac{0.5 \times 33.32}{0.8} = 20.8\,\text{N}$ (3 s.f.) (1)
M1 Energy to the top: $v^2 = 49 - 2 \times 9.8 \times 1.6 = 17.64$ (1)
M1 At the top: $T + 0.5 \times 9.8 = \dfrac{0.5 \times 17.64}{0.8}$ (1)
A1 $T = 6.125\,\text{N}$ (accept $6.13\,\text{N}$ ) (1)
B1 $T > 0$ at the top (the point of minimum tension), so the string stays taut and the circle is completed (1)