A certain wheel has a rotational inertia of 12 kg∙m².As it turns through 5.0 rev its angular velocity increases from 5.0 rad/s to 6.0 rad/s.If the net torque is constant its value is: A)0.015 N∙m B)0.18 N∙m C)0.57 N∙m D)2.1 N∙m E)13 N∙m

Question

Quizplus · Accepted Answer

We can use the equation for rotational kinetic energy:

$K_{rot}=\frac{1}{2}I\omega^2$

The initial kinetic energy is:

$K_{i}=\frac{1}{2}(12	ext{ kg}\cdot	ext{m}^2)\cdot (5.0	ext{ rad/s})^2=750	ext{ J}$

The final kinetic energy is:

$K_{f}=\frac{1}{2}(12	ext{ kg}\cdot	ext{m}^2)\cdot (6.0	ext{ rad/s})^2=1080	ext{ J}$

The change in kinetic energy is:

$\Delta K=K_{f}-K_{i}=330	ext{ J}$

We know that the net torque is constant, so we can use the equation for work done by torque:

$W_{net}=\Delta K=	au\Delta	heta$

We want the torque, so we can solve for it:

$	au=\frac{\Delta K}{\Delta	heta}=\frac{330	ext{ J}}{(5.0	ext{ rev})\cdot(2\pi	ext{ rad/rev})}=\boxed{2.1	ext{ N}\cdot	ext{m}}$

A Certain Wheel Has a Rotational Inertia of 12 Kg∙m2

A Certain Wheel Has a Rotational Inertia of 12 Kg∙m²