Question

A wheel has a constant angular acceleration of $3.0$ $rad/s^2$, During a certain $4.0s$ interval, it turns through an angle of $120rad$. Assuming that. $at=0$, angular speed ${\omega }_0=3rad/s$ how long is motion at the start of this $4.0second$ interval ?

• A. $7s$
• B. $9s$
• C. $4s$
• D. $10s$

Answer 1

The correct option is A $7s$

we have

angular acceleration $\alpha =3rad$ $s^{-2}$

and at $t=0$ ${\omega }_0=3rad$ $s^{-1}$

Let us say at some time $t$, the wheels turns an angle of ${\theta }_1$ and at $t+4$ it turns angle ${\theta }_2$

$\Rightarrow$ ${\theta }_1-{\theta }_2=120rad$

we know that

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

at time $t$

$\Rightarrow$ ${\theta }_1={\omega }_{0}t+\frac{1}{2}\alpha t^2$

i.e. ${\theta }_1=3t+\frac{3}{2}{t}^2......\left(i\right)$

and at $t+4$

${\theta }_2=3(t+4)+\frac{3}{2}{(t+4)}^2$

${\theta }_2=3t+12+\frac{3}{2}(t^2+16+3t)....\left(ii\right)$

Subtracting (i) from (ii)

${\theta }_2-{\theta }_1=[3t+12+\frac{3}{2}(t^2+16+8t\left)\right]-[3t+\frac{3}{2}{t}^2]$

$\Rightarrow$ $120=12+\frac{3}{2}[16+8t]$

$\Rightarrow$ $16+8t=\frac{2}{3}(120-12)$

$\Rightarrow$ $8t=72-16$

$\Rightarrow$ $t=\frac{56}{8}=7s$

Hence

At the start of $4s$ interval the motion is at $7$th second