Question

A pulley wheel of diameter $10\text{cm}$ has a $1\text{m}$ long cord wrapped around its periphery. The wheel is initially at rest. Assuming angular acceleration of wheel to be uniform and the time taken for the cord to unwind completely to be $18\text{s}$, find its tangential acceleration.

• A. $6.17{\text{mm/s}}^2$
• B. $2{\text{mm/s}}^2$
• C. $0.1{\text{mm/s}}^2$
• D. $0.01{\text{mm/s}}^2$

Answer 1

The correct option is A $6.17{\text{mm/s}}^2$

Given,
Initial angular velocity $\left({\omega }_o\right)=0\text{rad/s}$
Time taken to unwind completely $\left(t\right)=18\text{s}$

Length of cord wound on pulley $=1\text{m}$

Angle of windings, $\left(\theta \right)=\frac{l}{r}=\frac{1}{0.05}=20\text{rad}$

Since angular acceleration is uniform, from the equation
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2\Rightarrow \theta =\frac{1}{2}\alpha t^2$
($\because {\omega }_0=0$)
$\Rightarrow \alpha =\frac{2\theta }{t^2}=\frac{2\times 20}{(18{)}^2}{\text{rad/s}}^2$
Tangential acceleration $a_t=r\alpha$
$\Rightarrow a_t=0.05\times \frac{2\times 20}{(18{)}^2}{\text{m/s}}^2$
$\Rightarrow a_t=6.17\times {10}^{-3}{\text{m/s}}^2$
$\therefore a_t=6.17{\text{mm/s}}^2$