Question

(e) If the wheel rolls along the ground without slipping, the instantaneous velocity of the atoms of the object that are momentarily in contact with the ground is zero. This zero-velocity condition implies that , where is the angular speed of the object, since the instantaneous speed of the contact point is . During the time between 10 s and 15 s, how far did the center of the wheel move, in meters?

Answer 1

Answer:

distance moved by the center of the wheel during time $t_1 = 10 \ s$  to   $t_2 = 15\ s$  is  145.32 m

Explanation:

Given that :

The radius of the wheel  R =  28 cm

Initial angular speed of the wheel $\theta_o = 16 rev/s$

= $16*\frac{2}{1} *\frac{\pi}{1}rad/s$

= $32 \pi\ rad/s$

= 100.5 rad/s

At time t = 9 s ;  the angle  rotated by wheel $\theta_o = \omega_o t$

= 32 π × 9

= 905 rad

At time $t_1 = 10 \ s$; the angular speed is definitely the same and the initial velocity $\omega_1$ is $\omega_o = 32 \pi \ rads$

However ; after time  $t_1 = 10 \ s$ ; the angular acceleration of the wheel ∝ = 1.3 rad/s²

At time $t_2 = 15\ s$ ; angular speed of wheel $\omega_2 = \omega_1 + \alpha \delta \ t$

$\omega_2 = 32 \pi + 1.3( 15 - 10)$

$\omega_2 =107 \ rad/s$

Now; the angle rotated by the wheel during time  $t_1 = 10 \ s$  to   $t_2 = 15\ s$  is expressed as:

$\theta_1 = \omega_1(t_2-t_1)+ \frac{1}{2} \alpha (t_2-t_1)^2\\\\\theta_1 = 32 \pi (15-10) + \frac{1}{2}*1.3*(15-10)^2\\\\\theta_1 = 519 \ rad$

From the question; we are being told that :

This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R.

i.e $v_{cm} = R \omega$

then we can say :

acceleration $a_{cm} = R \alpha$    and

linear distance $s = R \theta$    indicating   how far did the center of the wheel move

So; distance moved by the center of the wheel during time $t_1 = 10 \ s$  to   $t_2 = 15\ s$  is ;

$s = R \theta$

= 0.28 × 519

s = 145.32 m

Therefore; distance moved by the center of the wheel during time $t_1 = 10 \ s$  to   $t_2 = 15\ s$  is  145.32 m