Question

A particle of mass m is projected at time $t=0$ with a velocity u making an angle ${45}^{\circ }$ with the horizontal. The magnitude of the torque due to weight of the particle is $\left(\frac{xm{u}^2}{4}\right)$, when it is at its maximum height, about a point where the particle was at time $t=\frac{u}{2\sqrt{2}g}$ on the trajectory. Find the vaule of x

Answer 1

At $t=\frac{u}{2\sqrt{2}g}$ we get the coordinates of the particle as
$s_x=ucos{45}^{o}t=\frac{u}{\sqrt{2}}\times \frac{u}{2\sqrt{2}g}=\frac{u^2}{4g}$
and similarly
$s_y=usin{45}^{o}t-\frac{1}{2}g{t}^2=\frac{3u^2}{16g}$
Thus the coordinates are $B(\frac{u^2}{4g},\frac{3u^2}{16g})$
Now the coordinates of the point A which is at its maximum height is given as $(\frac{R}{2},h)$
Using the formulas for maximum height $h=\frac{u^{2}si{n}^2\theta }{2g}$ and range of projectile $R=\frac{u^{2}sin2\theta }{g}$ we get the coordinates as
$(\frac{u^2}{2g},\frac{u^2}{4g})$
Now the torque is given as the gravitational force $mg$ times the horizontal distance between the two points.

Horizontal distance between the two points is $\frac{u^2}{2g}-\frac{u^2}{4g}=\frac{u^2}{4g}$.
Thus we get
$\tau =mg\times \frac{u^2}{4g}$
Solving this we get
$\tau =\frac{m{u}^2}{4}$
Thus we get $x$ as 1.