A cord of negligible mass is wound round the rim of a fly wheel of mass $20 \ kg$ and radius $20 \ cm$. A steady pull of $25 N$ is applied on the cord as shown in Fig. $6.31$. The flywheel is mounted on a horizontal axle with frictionless bearings.
$(a)$ Compute the angular acceleration of the wheel.
$(b)$ Find the work done by the pull, when $2m$ of the cord is unwound.
$(c)$ Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
$(d)$ Compare answers to parts $(b)$ and $(c).$
$(a)$ Compute the angular acceleration of the wheel.
$(b)$ Find the work done by the pull, when $2m$ of the cord is unwound.
$(c)$ Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
$(d)$ Compare answers to parts $(b)$ and $(c).$
Example-(6.12)
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$(a)\text { We use } I \alpha=\tau$
$\text { the torque } \tau=F R$
$ =25 \times 0.20 Nm (\text { as } R=0.20 m )$
$ =5.0 Nm$
$I=$ Moment of inertia of flywheel about its
$\text { axis }=\frac{M R^2}{2}$
$=\frac{20.0 \times(0.2)^2}{2}=0.4 \ kg m ^2$
$\alpha=\text { angular acceleration }$
$=5.0 N m / 0.4 \ kg m ^2=12.5 s ^{-2}$
$(b)$ Work done by the pull unwinding $2 m$ of the cord
$=25 N \times 2 m =50 J$
$(c)$ Let $\omega$ be the final angular velocity.
The kinetic energy gained $=\frac{1}{2} I \omega^2$, since the wheel starts from rest.
Now, $\omega^2=\omega_0^2+2 \alpha \theta, \omega_0=0$
The angular displacement $\theta=$ length of unwound string / radius of wheel $=2 m / 0.2 m =10 rad$
$\omega^2=2 \times 12.5 \times 10.0=250( rad / s )^2$
$\therefore \text { K.E.gained }=\frac{1}{2} \times 0.4 \times 250=50 J$
$(d)$ The answers are the same, i.e. the kinetic energy gained by the wheel $=$ work done by the force.
There is no loss of energy due to friction.
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