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 made to travel in a horizontal circle at constant speed, as shown in Fig. 2.2. The distance from the sphere’s centre to the top of the spring is now $10.8\,\text{cm}$.
(a)[3]
Explain, with reference to the forces on the sphere, why the spring in Fig. 2.2 is longer than the spring in Fig. 2.1.
(b(i))[1]
The angle between the spring’s line of action 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 has a value of $3.2\,\text{N}$.
(b(iii))[2]
The spring obeys Hooke’s law. Calculate the spring constant, in $\text{N cm}^{-1}$, for the spring.
(c(i))[2]
Use the information in (b) to determine the sphere’s centripetal acceleration.
(c(ii))[2]
Calculate the period for the sphere’s circular motion.
Worked solution & mark scheme
This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: “a horizontal force provides the centripetal acceleration” …