Question

A block of mass $12Kg$ is attached to a string wrapped around a wheel of radius $10cm$. The acceleration of the block moving down an inclined plane is measured at $2m/s^2$. The moment of inertia of the wheel is

• A. $0.28Kg{m}^2$
• B. $0.56Kg{m}^2$
• C. $0.92Kg{m}^2$
• D. $0.74Kg{m}^2$

Answer 1

The correct option is A $0.28Kg{m}^2$

From the FBD of the block we get the equation as
$masin\theta +T=mgsin\theta$
or
$T=m(g-a)sin\theta$
The torque balance at the pulley gives us
$Tr=I\alpha$
or
$Tr=I\frac{a}{r}$
Thus we get the moment of inertia as
$I=\frac{T{r}^2}{a}$
or
$I=\frac{m(g-a)sin\theta r^2}{a}$
or
$I=\frac{12(9.8-2)0.6\times 0.01}{2}=0.28kg{m}^2$