Question

Ball bearings leave the horizontal trough with a velocity of magnitude $u$ and fall through the $70mm$ diameter hole as shown. Calculate the permissible range of $u$ which will enable the balls to enter the hole. Take the dotted positions to represent the limiting conditions. (Take $g=10m/s^2\text{and}\sqrt{10}=3.2$)

• A. $u=0.62m/s$ to $1.01m/s$
• B. $u=0.76m/s$ to $1.16m/s$
• C. $u=0.43m/s$ to $1.91m/s$
• D. $u=0.80m/s$ to $1.03m/s$

Answer 1

The correct option is B $u=0.76m/s$ to $1.16m/s$



Horizontal distance travelled by the ball in time $t$,
$s=ut--\left(1\right)$
Vertical distance travelled by the ball in time $t$,
$h=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{2h}{g}}--\left(2\right)$
Substituting $\left(2\right)$ in $\left(1\right)$, $u=\frac{s}{\sqrt{\frac{2h}{g}}}$
To fall through the hole, the centre of the ball bearing must be within the range of C and D with limited trajectories between BC and BD.

For path BC, $S_{BC}=(120-\frac{D}{2})+\frac{d}{2}=(120-\frac{70}{2})+\frac{20}{2}=95mm$

$\Rightarrow u_{min}=u_{BC}=\frac{95\times {10}^{-3}}{\sqrt{\frac{2\times 80\times {10}^{-3}}{10}}}$
$=\frac{95\times {10}^{-3}}{\sqrt{\frac{16\times {10}^{-2}}{10}}}$
$=\frac{95\times {10}^{-3}}{0.4}\times 3.2=0.76m{s}^{-1}$

For path BD, $S_{BD}=(120+\frac{D}{2})-\frac{d}{2}=(120+\frac{70}{2})-\frac{20}{2}=145mm$

$\Rightarrow u_{max}=u_{BD}=\frac{145\times {10}^{-3}}{\sqrt{\frac{2\times 80\times {10}^{-3}}{10}}}$
$=\frac{145\times {10}^{-3}}{0.4}\times 3.2=1.16m{s}^{-1}$