91影视

Mass of the grinding wheel$\left(m\right)$= 2.80-kg.

Radius of the solid cylinder $\left(R\right)$= 0.100 m.

Angular speed $\left(蝇\right)$= 1200 rev/min.

In order to find the torque, we will use the equations $\left(9.5\right)$and $\left(10.7\right)$.

Therefore,

localid="1667974204468" ${\mathbf{伪}}_{\mathbf{z}}\mathbf{=}\frac{{\mathbf{d蝇}}_{\mathbf{z}}}{\mathbf{dt}}$

Substitute the values and solve:

$$\begin{gathered} 鈬�{伪}_z=\frac{1200-0}{2.5} \\ 鈬�{伪}_z=\frac{1200\frac{2蟺}{60}}{2.5} \\ 鈬�{伪}_z=50.24鈥�\frac{rad}{s^2} \end{gathered}$$

Now, solve for the torque as:

$$\begin{gathered} 蟿=I{伪}_z \\ 鈬�蟿=\frac{1}{2}m{R}^2{伪}_z \\ 鈬�蟿=\frac{1}{2}\left(2.80\right){\left(0.100\right)}^2\left(50.24\right) \\ 鈬�蟿=0.70鈥�Nm \end{gathered}$$

Hence, the torque is, $蟿=0.70鈥�Nm$.

Let us find the angle by using formula $\left(9.12\right)$.

Here,

${蝇}_0=0$

Thus,

${蝇}_z^2={蝇}_{0z}^2+2{伪}_z螖胃$

Substitute the values and solve as:

$$\begin{gathered} 鈬�{蝇}_z^2=0+2{伪}_z螖胃 \\ 鈬�螖胃=\frac{{蝇}_z^2}{2{伪}_z} \\ 鈬�螖胃=\frac{{\left(125.6\right)}^2}{2\left(50.24\right)} \\ 鈬�螖胃=157.4鈥�rad \end{gathered}$$

Hence, the angle is, $螖胃=157.4鈥�rad$

The work done is calculated by using the formula given by,

Hence, the work done by the torque is, $W=110.2J$.

The kinetic energy is given by,

$K=\frac{1}{2}I{蝇}^2$

Substitute the values and solve as:

Hence, the kinetic energy of the grinding wheel is, $K=110鈥�J$.