91Ó°ÊÓ

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

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

Angular speed $\left(\mathrm{Ó\neg }\right)$= 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" ${\mathbf{Î\pm }}_{\mathbf{z}}\mathbf{=}\frac{{\mathbf{dÓ\neg }}_{\mathbf{z}}}{\mathbf{dt}}$

Substitute the values and solve:

$$\begin{gathered} ⇒{\mathrm{Î\pm }}_z=\frac{1200-0}{2.5} \\ ⇒{\mathrm{Î\pm }}_z=\frac{1200\frac{2\mathrm{Ï€}}{60}}{2.5} \\ ⇒{\mathrm{Î\pm }}_z=50.24 \frac{rad}{s^2} \end{gathered}$$

Now, solve for the torque as:

$$\begin{gathered} \mathrm{Ï„}=I{\mathrm{Î\pm }}_z \\ ⇒\mathrm{Ï„}=\frac{1}{2}m{R}^2{\mathrm{Î\pm }}_z \\ ⇒\mathrm{Ï„}=\frac{1}{2}\left(2.80\right){\left(0.100\right)}^2\left(50.24\right) \\ ⇒\mathrm{Ï„}=0.70 Nm \end{gathered}$$

Hence, the torque is, $\mathrm{τ}=0.70 Nm$.

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

Here,

${\mathrm{Ó\neg }}_0=0$

Thus,

${\mathrm{Ó\neg }}_z^2={\mathrm{Ó\neg }}_{0z}^2+2{\mathrm{Î\pm }}_z\mathrm{Δ}\mathrm{θ}$

Substitute the values and solve as:

$$\begin{gathered} ⇒{\mathrm{Ó\neg }}_z^2=0+2{\mathrm{Î\pm }}_z\mathrm{Δ}\mathrm{θ} \\ ⇒\mathrm{Δ}\mathrm{θ}=\frac{{\mathrm{Ó\neg }}_z^2}{2{\mathrm{Î\pm }}_z} \\ ⇒\mathrm{Δ}\mathrm{θ}=\frac{{\left(125.6\right)}^2}{2\left(50.24\right)} \\ ⇒\mathrm{Δ}\mathrm{θ}=157.4 rad \end{gathered}$$

Hence, the angle is, $\mathrm{Δ}\mathrm{θ}=157.4 rad$

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,

$K=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

Substitute the values and solve as:

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