Question

A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter $1\text{m}$. A mass of $2\text{kg}$ is tied to the end of the cord and it s allowed to fall from rest. The weight falls $2\text{m}$ in $4\text{s}$. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. The angular acceleration produced is (Take $g=10{\text{ms}}^{-2}$)

• A. $0.5{\text{rad s}}^{-2}$
• B. $1.0{\text{rad s}}^{-2}$
• C. $2.0{\text{rad s}}^{-2}$
• D. $4.0{\text{rad s}}^{-2}$

Answer 1

The correct option is A $0.5{\text{rad s}}^{-2}$


We first find the acceleration $a$ by using the relation
$s=ut+\frac{1}{2}a{t}^2$
or $2=0+\frac{1}{2}\times a\times (4{)}^2$
which gives $a=\frac{1}{4}{\text{ms}}^{-2}$. Now $R=0.5\text{m}$.

We know that the relation between linear acceleration $\left(a\right)$ and angular acceleration $\left(\alpha \right)$ for pure rolling is
$\alpha =\frac{a}{R}=\frac{1}{4\times 0.5}=\frac{1}{2}=0.5{\text{rad s}}^{-2}$

Hence the correct choice is (a).