Problems

Filters
Clear Filters

1 problem found

2015 Paper 2 Q10
D: 1600.0 B: 1516.0
Newton's laws connected particles tension trigonometry acceleration constraint motion particle on floor normal reaction leaving surface

A particle of mass \(m\) is pulled along the floor of a room in a straight line by a light string which is pulled at constant speed \(V\) through a hole in the ceiling. The floor is smooth and horizontal, and the height of the room is \(h\). Find, in terms of \(V\) and \(\theta\), the speed of the particle when the string makes an angle of \(\theta\) with the vertical (and the particle is still in contact with the floor). Find also the acceleration, in terms of \(V\), \(h\) and \(\theta\). Find the tension in the string and hence show that the particle will leave the floor when \[ \tan^4\theta = \frac{V^2}{gh}\,. \]

Solution:

TikZ diagram
The length of the string is \(h/\cos \theta\), and it is decreasing at a rate \(V\). The distance along the ground is decreasing at a rate of \(V/\sin \theta\). Note that \(-V = \frac{\d}{\d t} \left ( \frac{h}{\cos \theta} \right) = \frac{h} {\cos^2 \theta} \sin \theta \cdot \dot{\theta} \Rightarrow \dot{\theta} = -\frac{V\cos^2\theta}{h \sin \theta}\). Note that \(a = \frac{\d}{\d t} \left ( \frac{V}{\sin \theta} \right) = -\frac{V}{\sin^2 \theta} \cos \theta \cdot \dot{\theta} = \frac{V^2 \cos^3 \theta}{h\sin^3 \theta}\). Resolving horizontally we must have \(T \sin \theta = ma \Rightarrow T = \frac{V^2m \cos^3 \theta}{h \sin^4 \theta}\). Resolving vertically at the point where we are about to leave the ground, we must have \(T\cos \theta = mg \Rightarrow \frac{V^2m \cos^4 \theta}{h \sin^4 \theta} = mg \Rightarrow \tan^4 \theta = \frac{V^2}{gh}\)

View