91影视

Angular speed is the rate at which the central angle of a spinning body varies over time.

The rate of change of angular displacement is known as angular speed.

Mass of wheel,$m=6kg$

Radius of the wheel,$r=0.3m$

As the wheel starting from rest, initial angular speed of the wheel,${蝇}_i=0rad/s$

Applied force at the edge of the wheel,$F=15N$

Total time that force applied at the edge of the wheel,$t=0.6s$

Torque acting on the system related to initial angular momentum and final angular momentum as

${\stackrel{鈫�}{蟿}}_{net}=\frac{L_f-L_i}{螖t}$

Here,

$L_f=I{蝇}_f鈫�$Final angular momentum of the wheel

$L_i=I{蝇}_i鈫�$Initial angular momentum of the wheel

$螖t鈫�$Time taken

${\stackrel{鈫�}{蟿}}_{net}鈫�$Net torque

Net torque acting on the wheel${\stackrel{鈫�}{蟿}}_{net}$ related to force acting at the edge$\left(F\right)$ and perpendicular distance between the axis of rotation and force as

${\stackrel{鈫�}{蟿}}_{net}=Fr$

From the above two relations we get

$Fr=\frac{I{蝇}_f-I{蝇}_i}{螖t}$

Isolating the final angular velocity of the wheel from above equation we get

${蝇}_f=\frac{Fr-螖t}{螖t}+{蝇}_i$

(a)Moment of inertia of the uniform density wheel is

$I=\frac{m{r}^2}{2}$

Now the final angular velocity of the wheel is

$\begin{array}{rcl}{蝇}_f& =& \frac{2Fr螖t}{m{r}^2}+0\\ & =& \frac{2Fr螖t}{m{r}^2}\\ & =& \frac{2F螖t}{mr}\\ & & \end{array}$ (Since ${蝇}_i=0rad/s$)

By substituting the given values in above equation, we get

$\begin{array}{rcl}{蝇}_i& =& \frac{2\left(15N\right)\left(0.6s\right)}{\left(6kg\right)\left(0.3m\right)}\\ & =& 10rad/s\\ & & \end{array}$

Final angular speed of the wheel is$10rad/s$ .

(b)Initial angular speed,${蝇}_i=0$

Final angular speed,${蝇}_f=10rad/s$

Now the average angular speed of the wheel is

$\begin{array}{rcl}{蝇}_{avg}& =& \frac{{蝇}_i+{蝇}_f}{2}\\ & =& \frac{0+10rad/s}{2}\\ & =& 5rad/s\\ & & \end{array}$

The average angular speed of the wheel is $5rad/s$.

(c)Angular displacement of the wheel is

$\begin{array}{rcl}螖胃& =& \left(\frac{{蝇}_i+{蝇}_f}{2}\right)螖t\\ & =& \left(5rad/s\right)\left(0.6s\right)\\ & =& 3.0rad\\ & & \end{array}$

Angle turn by the wheel in$0.6s$ is$3.0rad$.

(d) The length of the string that comes off the wheel related to angular displacement of the wheel as

$\begin{array}{rcl}螖x& =& r螖胃\\ & =& \left(0.3m\right)\left(3.0rad/s\right)\\ & =& 0.9m\\ & & \end{array}$

The total length of the string that comes off the wheel is$0.9m$ .