Moderate -0.8 This is a straightforward recall question on circular motion requiring only the standard horizontal force equation F = mrω². The student must recognize that the horizontal component T sin θ provides the centripetal force, making this a single-step selection from given options with no calculation or derivation needed.
A conical pendulum consists of a light string and a particle of mass \(m\) kg
The conical pendulum completes horizontal circles with radius \(r\) metres and angular speed \(\omega\) radians per second. The string makes an angle \(\theta\) with the downward vertical.
The tension in the string is \(T\) newtons.
The conical pendulum and the forces acting on the particle are shown in the diagram.
\includegraphics{figure_3}
Which one of the following statements is correct?
Tick (\(\checkmark\)) one box.
[1 mark]
\(T \cos \theta = mr\omega^2\) \quad \(\square\)
\(T \sin \theta = mr\omega^2\) \quad \(\square\)
\(T \cos \theta = \frac{m\omega^2}{r}\) \quad \(\square\)
\(T \sin \theta = \frac{m\omega^2}{r}\) \quad \(\square\)
A conical pendulum consists of a light string and a particle of mass $m$ kg
The conical pendulum completes horizontal circles with radius $r$ metres and angular speed $\omega$ radians per second. The string makes an angle $\theta$ with the downward vertical.
The tension in the string is $T$ newtons.
The conical pendulum and the forces acting on the particle are shown in the diagram.
\includegraphics{figure_3}
Which one of the following statements is correct?
Tick ($\checkmark$) one box.
[1 mark]
$T \cos \theta = mr\omega^2$ \quad $\square$
$T \sin \theta = mr\omega^2$ \quad $\square$
$T \cos \theta = \frac{m\omega^2}{r}$ \quad $\square$
$T \sin \theta = \frac{m\omega^2}{r}$ \quad $\square$
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q3 [1]}}