Question

The minimum tangential speed of a rotating cylinder, at which the floor below the man may be removed and the man hangs resting against the wall is: [Take $g=10{\text{m/s}}^2$]

• A. $5\text{m/s}$
• B. $6\text{m/s}$
• C. $10\text{m/s}$
• D. $8\text{m/s}$

Answer 1

The correct option is C $10\text{m/s}$


For minimum tangential speed$\left(v\right)$, the angular speed $\left(\omega \right)$ of rotation will also be minimum.
On applying equilibrium condition for the man along vertical direction,
$f-mg=0$
$\Rightarrow f=mg$
$\Rightarrow \mu N=mg...\left(1\right)$
[limiting case for minimum angular velocity]
On applying equation of dynamics towards the centre of horizontal circular path,
$N=ma$
where $a$ is the centripetal acceleration, i.e $a=\frac{v^2}{r}$
$\Rightarrow N=\frac{m{v}^2}{r}...\left(2\right)$
Substituting Eq.$\left(2\right)$ in Eq.$\left(1\right)$,
$\frac{\mu m{v}^2}{r}=mg$
$\Rightarrow v=\sqrt{\frac{gr}{\mu }}$
$\Rightarrow v=\sqrt{\frac{10\times 2}{0.2}}$
[$\because r=2\text{m}$ and $\mu =0.2$ given]
$\therefore v=10\text{m/s}$
Minimum speed at which the floor may be removed is $10\text{m/s}$.