Question

If a constant torque of 500 Nm turns a wheel of
moment of inertia 100 kg m about an axis
through its centre, find the gain in angular veloc-
ity in 2s.​

Answer 1

AnswEr :

Gain in angular velocity in 2s is 10 rad/sec.

Explanation :

We're given with the constant torque and a wheel with it's moment of inertia about an axis through its centre. So, we need to find the gain in angular velocity in 2s.

• $\bf \tau$ = 500 Nm

• Moment of inertia (I) = 100 kg m²

• Time (t) = 2s

Now, we need to find the value of gain in angular velocity, that is,

• $\bf \omega_1 - \omega_2 = ?$

Also we know that, torque takes the place of force in linear kinematics.

So, direct equivalent to Newton's 2ⁿᵈ law of motion which is given by,

$\: \: \star \: { \boxed{\sf{ \purple {\tau = i \times \alpha}}}}$

Where, τ is torque produced, I is moment of inertia and α is given by the change in angular acceleration.

Substituting the values,

$:\implies \sf \: \tau = i \times \alpha \\ \\ \\ :\implies \sf \: \tau = i \times \dfrac{ \omega_{1} - \omega_{2}}{t} \\ \\ \\ :\implies \sf \: \omega_{1} - \omega_{2} = \dfrac{ \tau \times t}{i} \\ \\ \\ :\implies \sf \: \omega_{1} - \omega_{2} = \dfrac{ 500 \times 2}{100} \\ \\ \\ :\implies \sf \: \omega_{1} - \omega_{2} = \cancel{\dfrac{ 1000}{100}} \\ \\ \\ :\implies \sf\: { \underline{ \boxed{ \frak{ \pink{\omega_{1} - \omega_{2} = 10 \: rad \: sec{}^{ - 1} }}}}}$

$\therefore$ Gain in angular velocity in 2s is 10 rad/sec .

Answer 2

Explanation:

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