Answer:
distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex] to [tex]t_2 = 15\ s[/tex] is 145.32 m
Explanation:
Given that :
The radius of the wheel R = 28 cm
Initial angular speed of the wheel [tex]\theta_o = 16 rev/s[/tex]
= [tex]16*\frac{2}{1} *\frac{\pi}{1}rad/s[/tex]
= [tex]32 \pi\ rad/s[/tex]
= 100.5 rad/s
At time t = 9 s ; the angle rotated by wheel [tex]\theta_o = \omega_o t[/tex]
= 32 π × 9
= 905 rad
At time [tex]t_1 = 10 \ s[/tex]; the angular speed is definitely the same and the initial velocity [tex]\omega_1[/tex] is [tex]\omega_o = 32 \pi \ rads[/tex]
However ; after time [tex]t_1 = 10 \ s[/tex] ; the angular acceleration of the wheel ∝ = 1.3 rad/s²
At time [tex]t_2 = 15\ s[/tex] ; angular speed of wheel [tex]\omega_2 = \omega_1 + \alpha \delta \ t[/tex]
[tex]\omega_2 = 32 \pi + 1.3( 15 - 10)[/tex]
[tex]\omega_2 =107 \ rad/s[/tex]
Now; the angle rotated by the wheel during time [tex]t_1 = 10 \ s[/tex] to [tex]t_2 = 15\ s[/tex] is expressed as:
[tex]\theta_1 = \omega_1(t_2-t_1)+ \frac{1}{2} \alpha (t_2-t_1)^2\\\\\theta_1 = 32 \pi (15-10) + \frac{1}{2}*1.3*(15-10)^2\\\\\theta_1 = 519 \ rad[/tex]
From the question; we are being told that :
This zero-velocity condition implies that vCM=ωR, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ωR.
i.e [tex]v_{cm} = R \omega[/tex]
then we can say :
acceleration [tex]a_{cm} = R \alpha[/tex] and
linear distance [tex]s = R \theta[/tex] indicating how far did the center of the wheel move
So; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex] to [tex]t_2 = 15\ s[/tex] is ;
[tex]s = R \theta[/tex]
= 0.28 × 519
s = 145.32 m
Therefore; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex] to [tex]t_2 = 15\ s[/tex] is 145.32 m
