91影视

Mass of parachutists after landing is $M=\text{70.0}\text{kg}$.

Mass of table is $m=\text{120}\text{kg}$.

The radius of the table is $r=\text{2.00}\text{m}$.

The angular velocity of the table is $蝇=3\text{rad}/\text{s}$.

The law of conservation of angular momentum means that the angular momentum of a rotating object does not change unless some type of external torque is applied to it.

The moment of inertia of the turntable calculated as follows:

$$\begin{gathered} I_i=m{r}^2 \\ =\text{120}\text{kg}脳{\left(\text{2}\text{m}\right)}^{\text{2}} \\ =\text{480}{\text{kg.m}}^2 \end{gathered}$$

The moment of inertia after parachutist is solved as:

$$\begin{gathered} I_f=\left(M+m\right)r^2 \\ =\left(\text{120+70}\right)\text{kg}脳{\left(\text{2}\text{m}\right)}^{\text{2}} \\ =\text{760}{\text{kg.m}}^{\text{2}} \end{gathered}$$

Use the conservation of angular momentum to obtain angular speed:

$$\begin{gathered} {\text{I}}_i{蝇}_i=I_f{蝇}_f \\ \text{480}{\text{kg.m}}^{\text{2}}脳\text{3}\text{rad}/\text{s}=\left(\text{760}{\text{kg.m}}^{\text{2}}\right){蝇}_f \\ {蝇}_f=\frac{\text{480}{\text{kg.m}}^{\text{2}}脳\text{3}\text{rad}/\text{s}}{\text{760}{\text{kg.m}}^{\text{2}}} \\ {蝇}_f=\text{1.89}\text{rad}/\text{s} \end{gathered}$$

Hence, the final angular speed is $\text{1.89}\text{rad/s}$.

The speed of the turntable after the landing is calculated as follows:

$$\begin{gathered} v_i=r蝇 \\ =\text{2}\text{m}脳\text{1.89}\text{rad/s} \\ =\text{3.78}\text{m/s} \end{gathered}$$

The speed of the turntable before the landing is calculated as follows:

$$\begin{gathered} v_i=r蝇 \\ =\text{2}\text{m}脳\text{3.0}\text{rad/s} \\ =\text{6.0}\text{m/s} \end{gathered}$$

The kinetic energy after parachutist lands:

$$\begin{gathered} K{E}_f=\frac{\text{1}}{\text{2}}m{v}_f^{\text{2}} \\ =\frac{\text{1}}{\text{2}}\left(\text{190}\text{kg}\right){\left(\text{3.78}\text{m}/\text{s}\right)}^{\text{2}} \\ =\text{1357.39}\text{J} \end{gathered}$$

Thus, the kinetic energy after parachutist lands is $\text{1357.39}\text{J}$.

The kinetic energy before the parachutist lands is

$$\begin{gathered} K{E}_i=\frac{\text{1}}{\text{2}}m{v}_i^{\text{2}} \\ =\frac{\text{1}}{\text{2}}\left(\text{120}\text{kg}\right){\left(\text{6}\text{m}/\text{s}\right)}^{\text{2}} \\ =\text{2160}\text{J} \end{gathered}$$

Thus, the kinetic energy after parachutist lands is $\text{2160}\text{J}$.

Hence, the kinetic energy before parachutist lands and after parachutist land is not same because the angular speed changes.