Question

A small block of mass $4\text{kg}$ is attached to a cord passing through a hole in a horizontal frictionless surface as shown in figure. The block is originally revolving in a circle of radius $r=0.5\text{m}$ about a vertical axis passing through the hole, with a tangential speed of $v_o=4$ m/s. The cord is then pulled slowly from below, shortening the radius of the circle to $r=0.2\text{m}$, in which the block revolves. Find the new tangential speed of block. (Neglect the mass of cord)

• A. $5\text{m/s}$
• B. $10\text{m/s}$
• C. $20\text{m/s}$
• D. $25\text{m/s}$
  

Answer 1

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

The only external force acting on this system of $\text{(block+string)}$ is the weight of block ($mg=40\text{N}$). Hence ${\tau }_{\text{ext}}=0$, hence $PCAM$ can be applied for system.
Initial angular speed of block,
${\omega }_o=\frac{v_o}{r_o}=\frac{4}{0.5}=8\text{rad/s}$
$\therefore$Initial moment of inertia of block
$I_o=m{r}_o^2=4\times {0.5}^2{\text{kg.m}}^2$
$=1{\text{kg.m}}^2$
Final moment of inertia of block
$I=m{r}^2=4\times {0.2}^2=0.16{\text{kg.m}}^2$

Let final angular speed of block is $\omega \text{rad /s}$
$\therefore$ Applying $PCAM$ about the axis of rotation gives:

$L_{final}=L_{initial}$
$I\omega =I_o{\omega }_o$
$\Rightarrow \omega =\frac{I_o{\omega }_o}{I}=\frac{1\times 8}{0.16}=50\text{rad/s}$
Again using $v=\omega \times r,$ where $v$ is final tangential speed of the block.
$v=(50\times 0.2)\text{m/s}$
$v=10\text{m/s}$
$\therefore$ option $\left(b\right)$ is correct.