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$$\begin{array}{l}{\mathbf{\tau}}_{\mathrm{net}}=I\alpha \end{array}$$
Combining, two equations we have two expressions for torque :
Problem : A wheel of mass 10 kg and radius 0.2 m is rotating at an angular speed of 100 rpm with the help of a motor drive. At a particular moment, the motor is turned off. Neglecting friction at the axle, find the constant tangential force required to stop the wheel in 10 revolutions.
Solution : We need to find the tangential force, which is related to torque as :
$$\begin{array}{l}\tau =rF\end{array}$$
If we know torque, then we can find the force required to stop the wheel. Now, torque is given as (using Newton's second law in angular form),
$$\begin{array}{l}\tau =I\alpha \end{array}$$
Here, MI of the rotating wheel is :
$$\begin{array}{l}I=\frac{1}{2}M{R}^{2}=\frac{1}{2}10\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{\mathrm{0.2}}^{2}=\mathrm{0.2}\phantom{\rule{2pt}{0ex}}\mathrm{kg-}{m}^{2}\end{array}$$
By inspecting equation of torque, we realize that we need to find angular acceleration to calculate torque on the wheel. Since the external force is constant, we can conclude that the resulting torque is also constant. This means that the wheel is decelerated at constant rate. Thus, applying equation of motion,
$$\begin{array}{l}{\omega}_{2}^{2}={\omega}_{1}^{2}-2\alpha \theta \end{array}$$
Here,
$$\begin{array}{l}{\omega}_{1}=\frac{100x2\pi}{60}=\frac{10\pi}{3}\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s,\phantom{\rule{2pt}{0ex}}{\omega}_{2}=0,\phantom{\rule{2pt}{0ex}}\theta =10\phantom{\rule{2pt}{0ex}}\mathrm{rev}=2\pi \phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}10=20\pi \phantom{\rule{2pt}{0ex}}\mathrm{rad}\end{array}$$
$$\begin{array}{l}\Rightarrow \alpha =-\frac{{\omega}_{2}^{2}}{2\theta}=-\frac{{\left(\frac{10\pi}{3}\right)}^{2}}{2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}20\pi}=-\frac{5\pi}{18}\phantom{\rule{2pt}{0ex}}\mathrm{rad}/{s}^{2}\end{array}$$
Putting this value in the equation, we have :
$$\begin{array}{l}\Rightarrow \tau =I\alpha =0.2x-\frac{5\pi}{18}=-\frac{\pi}{18}\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
Thus,
$$\begin{array}{l}F=\frac{\tau}{r}=-\frac{\pi}{18x0.2}=-0.87\phantom{\rule{2pt}{0ex}}N\end{array}$$
Negative sign indicates that force is opposite to the direction of velocity.
Answer to this question is very relevant in the case of combined (translation and rotation) motion. Consider the case of rolling. The axis of rotation i.e. the reference of motion itself may be accelerating down an incline. We may also consider the example of a spring board diver. In this case, the diver accelerates down the spring board under gravity while performing eye catching twists and turns. Here, also the reference frame of rotation attached to the diver is accelerating.
Now, let us have the look the way we define a torque :
$$\begin{array}{l}\mathbf{t}=\mathbf{r}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}\mathbf{F}\end{array}$$
where,
$$\begin{array}{l}\mathbf{F}=m\mathbf{a}\end{array}$$
We ofcourse know that the relation of force and acceleration (which is Newton’s second law of motion) is valid in inertial frame and is not valid for non-inertial frame. We may, therefore, infer that the Newton’s second law in angular forms as derived here for system of particles and rigid body may also be not valid for non-inertial frames.
The question is, then, how do we apply these equation forms in cases of rolling and spring board driver ? The answer is that accelerated rolling or combined motion provides an unique situation for dealing motion in accelerated frame of reference.
If we recall, then we realize that we can render a non-inertial (accelerating reference) into an inertial frame by applying a psuedo force at the center of mass of the system of particle or the rigid body as the case may be. This has important bearing on the outcome of net torque. If the reference point is the center of mass of the system of particle, then the torque due to pseudo force about center of mass is zero. It means that applying Newton's law to a system of particle about the center of mass will be valid even if center of mass i.e. reference point itself is accelerating. Similar is the situation for rotation of a rigid body about an accelerating axis of rotation, which is passing through the center of mass. The torque due to pseudo force about the axis passing through center of mass is zero. Therefore, application of Newton's law for rotation of a rigid body about an accelerating axis passing through center of mass is valid.
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