Question

A block of mass $\mathrm{m}$ is attached to a pulley disc of equal mass $\mathrm{m}$ and radius $\mathrm{r}$ by means of a slack string as shown. The pulley is hinged about its center on a horizontal table and the block is projected with an initial velocity of $5 \mathrm{~m} / \mathrm{~s}$. Its velocity when the string becomes taut will be

A.$3~\text{m}/~\text{s}$
B.$2.5~\text{m}/~\text{s}$
C.$\dfrac{5}{3}~\text{m}/~\text{s}$
D.$\dfrac{10}{3} \mathrm{~m} / \mathrm{~s}$

Answer 1

In the center, a pulley is hinged and a massive thread is wrapped around it. Angular acceleration and its angular velocity rise will be due to torque supplied by force pulley. The hinge force is always equal and opposite to F. Calculate Force and momentum by using the theorem of momentum .

Formula used:
$F=\dfrac{\Delta(P)}{\Delta(t)}=\dfrac{m\left(v_{2}-v_{1}\right)}{t}$

Complete solution Step-by-Step:
The theorem of momentum variation applied to mass on the table says:
$F=\dfrac{\Delta(P)}{\Delta(t)}=\dfrac{m\left(v_{2}-v_{1}\right)}{t}$
where $v_{2}$ is the final speed and
$v_{1}$ the initial speed of the mass $m$
For the pulley the theorem of angular momentum variation says:
$M=\dfrac{\Delta(L)}{\Delta(t)}=\dfrac{I \cdot \omega_{2}}{t}$
where $M=(-F) \cdot r$
(here the minus sign shows the force on the pulley is opposing to the force on the mass on the table)
Therefore, we have two equations:
$(-F) \cdot r=\dfrac{I \cdot \omega_{2}}{t}$
$F=\dfrac{m\left(v_{2}-v_{1}\right)}{t}$
and we obtain
$\Rightarrow m\left( {{v}_{1}}-{{v}_{2}} \right)\cdot r=I\cdot {{\omega }_{2}}$ ------------ (1)
Now we can write for the angular acceleration of the pulley and its moment of inertia (the pulley is considered a solid cylinder)
$\omega_{2}=\dfrac{v_{2}}{r}$
and
$I=\dfrac{m \cdot r^{2}}{2}$
Replacing these two relations in (1) we get
$\Rightarrow m\left( {{v}_{1}}-{{v}_{2}} \right)\cdot r=\dfrac{m\cdot {{r}^{2}}}{2}\cdot \dfrac{{{v}_{2}}}{r}$
$2\left(v_{1}-v_{2}\right)=v_{2}$
$2 v_{1}=3 v_{2}$
${{v}_{2}}=\left( \dfrac{2}{3} \right)\cdot {{v}_{1}}=\dfrac{2}{3}\cdot 5=\dfrac{10}{3}=3.33m/s$

$\therefore$ The velocity of the mass on the table when the string is taut will be ${{v}_{2}}=\dfrac{10}{3}m/s$.

Note:
With a constant force F starting from resting, the thread is pulled. Its angular velocity remains the same as time increases, but the force on the hinge increases. At the Centre, a pulley is hinged and a massive thread is wrapped around it As time increases, the thread is pulled with a constant force F starting from rest.