Respuesta :
First we find the angular acceleration [tex] \alpha [/tex]. The angular velocities are [tex] \frac{240}{60} =4 \;rps [/tex] and [tex] \frac{360}{60} =6 \;rps [/tex] The angular velocities and [tex] \alpha [/tex] are related as
[tex] \omega=\omega_0+\alpha t\\
\alpha =\frac{\omega - \omega_0}{t} \\
\alpha =\frac{6-4}{(6.5)} \\
\alpha =0.3077 \;rps^2\\
\alpha =(0.3077 )2\pi =1.933 \;rad/s^2 [/tex]
Angle turned in 6.5 seconds is
[tex] 2 \alpha\theta = \omega^2-\omega_0^2\\
\theta = \frac{ \omega^2-\omega_0^2}{2 \alpha}\\
\theta = \frac{ 6^2-4^2}{2 (1.933)}\\
\theta = 5.173 \; rad [/tex]
The distance traveled by a point on the edge of the wheel is [tex] r \theta = \frac{33}{2}(5.173)= 85.35 \;cm [/tex]
The distance will a point on the edge of the wheel have traveled in 6.5 seconds of time is 33.7 meters.
How to find angular velocity of a body?
The angular velocity of a body is the rate by which the body changed its angle with respect to the time. It can be given as,
[tex]\omega= \dfrac{\Delta \theta}{\Delta t}[/tex]
A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. Convert it into the rad/s as,
[tex]\omega_o=240(0.1047)\\\omega=25.1\rm \; rad/s[/tex]
[tex]\alpha_2=360(0.1047)\\\alpha_2=37.7\rm \; rad/s[/tex]
The angular acceleration is rate of change of angular speed with time. It can be given as,
[tex]a=\dfrac{\Delta\omega}{t}\\a=\dfrac{\omega-\omega_o}{t}\\[/tex]
Put the values,
[tex]a=\dfrac{37.7-25.1}{6.5}\\a=1.94\rm\;rad/s^2[/tex]
By the equation of motion, the angle can be given as,
[tex]\theta=\omega_ot+\dfrac{1}{2}at^2\\\theta=25.1(6.5)+(0.5)(1.94)(6.5)^2\\\theta=204\rm \; rad[/tex]
The diameter of the wheel is 33 cm. Thus, the length of travel can be given as,
[tex]l=\theta\dfrac{d}{2}\\l=(204)\dfrac{0.33}{2}\\l=33.7\rm \; m[/tex]
Thus, the distance will a point on the edge of the wheel have traveled in 6.5 seconds of time is 33.7 meters.
Learn more about the angular velocity here;
https://brainly.com/question/1452612
