91影视

The rotational equivalent of force, is known as torque. It is also known as 鈥榤oment of force鈥� or 鈥榯urning effect of force鈥�. It plays the same role in rotational motion, as force plays in linear motion. It is a vector quantity, described as the cross product of force vector with the displacement vector.
$\stackrel{鈫�}{蟿}=\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{脳}\stackrel{\mathbf{鈫�}}{\mathbf{r}}$

The power$\left(\mathbf{P}\right)$ produced due to torque $\left(蟿\right)$in a motor rotating with angular velocity $\left(\mathbf{蝇}\right)$, is given as-

$$\begin{gathered} \mathbf{P}\mathbf{=}蟿\mathbf{脳}\mathbf{蝇} \\ \mathbf{蟿}\mathbf{=}\frac{\mathbf{P}}{\mathbf{蝇}} \end{gathered}$$

The angular speed in rad/s, is -

$$\begin{gathered} 蝇=4000rev/min\frac{2蟺rad}{60s}脳\frac{1min}{1rev} \\ =419\text{rad/s} \end{gathered}$$

Power output is-

$$\begin{gathered} P=150\text{kW} \\ \text{=150}脳{\text{10}}^3\text{W} \end{gathered}$$

The diameter of drum is-

$d=0.400\text{m}$

The torque is solved as follows:

$$\begin{gathered} P=蟿路蝇 \\ 蟿=\frac{P}{蝇} \\ 蟿=358\text{Nm} \end{gathered}$$

Then, the torque is $358\text{Nm}$.

On hanging the weight:

$$\begin{gathered} T-mg=0 \\ T=mg \end{gathered}$$

Weight on the drum is-

$w=mg$

For $T=mg$, torque is given as-

$$\begin{gathered} 蟿=T脳R \\ =mgR \end{gathered}$$

Solve for $m$as follows:

$$\begin{gathered} m=\frac{蟿}{gR} \\ =\frac{358\text{N.m}}{9.8\text{m}/{\text{s}}^{\text{2}}脳0.200\text{m}} \\ =73\text{kg} \end{gathered}$$

Thus, the weight that the motor can lift at constant speed is $73\text{kg}$.

The speed with which weight is rising can be calculated by-

$$\begin{gathered} P=T脳v \\ v=\frac{P}{T} \end{gathered}$$

For $P=150\text{kW}$, $T=mg$and $m=73\text{kg}$, we get-

$$\begin{gathered} v=\frac{150脳{10}^3\text{W}}{73\text{kg}脳\text{9.8}{\text{m}/\text{s}}^{\text{2}}} \\ v=209.7\text{m}/\text{s} \end{gathered}$$

Thus, the constant speed of the weight rise by$209.7鈥�\text{m}/\text{s}$.