Question

6) A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of
5 rad / sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm
for 10 seconds. Calculate the change in its angular momentum and change in its
kinetic energy
(Ans: 0.1kgm-/s,0.625 J)​

Answer 1

Given:

A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of 5 rad / sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm for 10 seconds.

To find:

• Change in angular momentum?
• Change in kinetic energy?

Calculation:

Torque is defined as rate of change of angular momentum:

$\therefore \: \tau = \dfrac{\Delta L}{\Delta t}$

$\implies \: \Delta L = \tau \times \Delta t$

$\implies \: \Delta L = 0.01 \times 10$

$\boxed{ \implies \: \Delta L = 0.1 \: kgm {s}^{ - 1} }$

New angular velocity :

$\therefore \: \omega = 5 + \alpha t$

$\implies \: \omega = 5 +( \dfrac{0.01}{ \frac{m {r}^{2} }{2} } \times 10)$

$\implies \: \omega = 5 +( \dfrac{0.01}{ \frac{8 \times {0.1}^{2} }{2} } \times 10)$

$\implies \: \omega = 5 +( \dfrac{1}{4} \times 10)$

$\implies \: \omega = 7.5 \: rad \: {s}^{ - 1}$

Now, change in KE :

$\therefore \: \Delta KE = \dfrac{1}{2} \times I \times \Delta( {\omega}^{2} )$

$\implies \: \Delta KE = \dfrac{1}{2} \times \dfrac{8 \times {0.1}^{2} }{2} \times ( {7.5}^{2} - {5}^{2} )$

$\implies \: \Delta KE = 0.02\times ( {7.5}^{2} - {5}^{2} )$

$\implies \: \Delta KE = 0.02\times 31.25$

$\boxed{ \implies \: \Delta KE = 0.625 \: joule}$

Hope It Helps.