Question

A small block is placed on a turn table. The coefficient of static friction between the block and the table is ${\mu }_s=0.5$. If the turn table is rotating with constant angular acceleration $\alpha =3{\text{rad/s}}^2$, what will be the angular speed at which the block begins to slip on the table ?
Given $r=1\text{m},g=10{\text{m/s}}^2.$

• A. $1\text{rad/s}$
• B. $2\text{rad/s}$
• C. $3\text{rad/s}$
• D. $4\text{rad/s}$

Answer 1

The correct option is B $2\text{rad/s}$

Since the turn table is rotating with constant acceleration $\alpha =3{\text{rad/s}}^2$, so at any instant, the block (rotating with table) will have two components of acceleration:

Tangential acceleration: $a_t=r\alpha$

Centripetal acceleration: $a_c={\omega }^{2}r$

So, total acceleration $a$ of the body will be:
$a=\sqrt{a_t^2+a_c^2}$

F.B.D of block:

Vertical equilibrium gives $N=mg$
Applying the condition of just slipping, friction will act at limiting value
$f_{max}={\mu }_{s}N={\mu }_{s}mg$
and friction will act in a direction opposite to that of total acceleration $a$

$ma=$ Pseudoforce on block, corresponding to total acceleration $a$
[ from frame attached with the body ]

From FBD, applying the condition of equilibrium when the block is just about to slip.

$f_{max}=ma$
$\Rightarrow {\mu }_{s}mg=m\sqrt{a_t^2+a_c^2}$
$\Rightarrow {\mu }_{s}g=\sqrt{(r\alpha {)}^2+({\omega }^{2}r{)}^2}$

Putting $\alpha =3{\text{rad/s}}^2,r=1\text{m},{\mu }_s=0.5$
$0.5\times 10=\sqrt{9+{\omega }^4}$
${\omega }^4=25-9=16$
$\therefore \omega =2\text{rad/s}$