Question

A chord of negligible mass is wound round the rim of fly wheel of mass 20kg and radius 20cm. A steady pull of 25N is applied on the cord. The fly is mounted on a horizontal axle with frictionless bearings. Find a)Angular acceleration of the wheel b) Work done by the pull when 2m of the cord is unwound c)Assuming the wheel starting from rest what is the gain in K.E.

Answer 1

here is your answer

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Answer 2

Answer: $\alpha = 12.5 rad/s^{2}$, $W = 50 J$ and $k = 50 J$

Explanation:

Given that,

Mass of the wheel M = 20 kg

Radius of the wheel R = 20 cm

Force F = 25 N

Length l = 2 m

We know that,

The torque is

$\tau = I \alpha$

$F\times R = \dfrac{MR^{2}}{2}\alpha$

(a). The angular acceleration of the wheel is

$\alpha = \dfrac{2F}{MR}$

$\alpha = \dfrac{2\times25 N}{20kg \times0.2\ m}$

$\alpha = 12.5 rad/s^{2}$

(b). Work done by the pull when 2 m of the cord unwound

$W = F\times d$

$W = 25\ N\times2m$

$W = 50 J$

(c). The gain in kinetic energy is

$k = \dfrac{1}{2}I\omega^{2}$

$k = \dfrac{1}{2}\times\dfrac{MR^{2}}{2}\times\omega^{2}$.....(I)

We know that,

The angular equation of motion is

$\omega^{2} = \omega_{0}^{2} + 2\alpha \theta$

Where,

$\omega$ = Final velocity of wheel

$\omega_{0}$ = Initial velocity

The wheel starting from rest so, the initial velocity will zero.

So,

$\omega^{2} = 2\alpha\theta$....(II)

We know,

$\theta = \dfrac{l}{r}$

$\theta = \dfrac{2}{0.2} = 10 radian$

Now, put the value of $\theta$ in equation (II)

$\omega^{2} = 2\times12.5\ rad/s^{2}\times10\ rad$

$\omega^{2}=250 (rad/sec)^{2}$

Now, put the value of $\omega^{2}$ in equation (I)

$k = \dfrac{1}{2}\times \dfrac{20\ kg\times 0.2m\times0.2m}{2}\times250\(rad/sec)^{2}$

$k = 50 J$

So, the gain in K.E is 50 J.

Hence, this is the required solution.