Physics 9702 · AS & A Level · Centripetal acceleration

Centripetal acceleration — practice question

A steel sphere with mass $0.29\,\text{kg}$ hangs in equilibrium from a vertical spring. In Fig. 2.1, the sphere’s centre is $8.5\,\text{cm}$ below the top of the spring. The sphere is then set into motion so that it travels round a horizontal circle at constant speed, as shown in Fig. 2.2. The distance from the centre of the sphere to the top of the spring is now $10.8\,\text{cm}$.
(a)[3]

Explain, with reference to the forces acting on the sphere, why the spring is longer in Fig. 2.2 than in Fig. 2.1.

(b(i))[1]

The angle between the linear axis of the spring and the vertical is $27^\circ$. Show that the radius $r$ of the circle is $4.9\,\text{cm}$.

(b(ii))[2]

Show that the tension in the spring is $3.2\,\text{N}$.

(b(iii))[2]

The spring obeys Hooke’s law. Calculate the spring constant, in $\text{N cm}^{-1}$, of the spring.

(c(i))[2]

Use the values from (b) to determine the centripetal acceleration of the sphere.

(c(ii))[2]

Calculate the period of the sphere’s circular motion.

(b)

The angle between the linear axis of the spring and the vertical measures $27^{\circ}$.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: The horizontal force provides the centripetal acceleration.

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