Shooting an arrow through the spokes of a turning wheel

[Ursa]

Homework Statement

The wheel in the figure has eight equally spaced spokes and a radius of 22 cm. It is mounted on a fixed axle and is spinning at 3.5 rev/s. You want to shoot a 26-cm-long arrow parallel to this axle and through the wheel without hitting any of the spokes. Assume that the arrow and the spokes are very thin. What minimum speed must the arrow have?

Relevant Equations

\theta - \theta_0 = \frac {1}{ 2} ( \omega_0 +\omega) t
x-x_0 = \frac {1}{2} (v_0+v)t

r=22 cm = 0.022 m 3.5 rev/s L_arrow = 26 cm= 0.026 m

first I got the speed in rad $$ 3.5* \frac {2 \pi}{0.022}= 999.6 m/s $$
from there I tried to determine the time the arrow had to pass through the spokes, 1/8 th of the wheel.
$$ \frac {2 \pi}{0.022} * \frac {1}{0.8} = 35.7 rad $$
inputting these in $\theta - \theta_0 = \frac {1}{ 2} ( \omega_0 +\omega) t$
$$ 35.7 = \frac {1}{2} (999.6+999.6)t $$
$$ t= \frac {35.7} {999.6} = 0.0357 s$$
then I took the length of the arrow as \Delta x
$$\Delta x = \frac {1}{2} (v_0+v)t$$
$$0.026 = \frac {1}{2} (v_0+v)* 0.0357$$
$$v= \frac {0.026} {\frac {1}{2} 0.0357} = 1.456 $$

now that I have written it down I think I forgot that the 1.456 m/s was (v_0 +v) and had it as my final answer v.
so would dividing that by 2 be the correct velocity of the arrow? 0.728 m/s?

 

[Orodruin]

You are making things hard on yourself by doing a lot of unnecessary intermediate computations.

Assume that the arrow enters the wheel just behind a spoke. How much would the wheel need to turn before the arrow is hit by the second spoke?

 

[Ursa]

You are making things hard on yourself by doing a lot of unnecessary intermediate computations.

Assume that the arrow enters the wheel just behind a spoke. How much would the wheel need to turn before the arrow is hit by the second spoke?

it would turn 1/8th of a rotation

I did $\frac {1}{3.5} * \frac {1}{8}$ before and also got $0.0357 s$ for the arrow to pass through.
I did the difficult version because I thought I may have done it wrong the first time when I did it this way.

 

[Orodruin]

it would turn 1/8th of a rotation

I did $\frac {1}{3.5} * \frac {1}{8}$ before and also got $0.0357 s$ for the arrow to pass through.
I did the difficult version because I thought I may have done it wrong the first time when I did it this way.

To be perfectly frank, your approach in the OP just looks weird and confused to me. For example, it is unclear to me what your 35.7 rad is. It is certainly not the rotation needed for the wheel to turn between spokes (which must be an eighth of 2pi, which is pi/4 < 1 rad).

it would turn 1/8th of a rotation

I did 13.5∗18 before and also got 0.0357s for the arrow to pass through.

This is correct. So, given that the arrow is 26 cm long, how fast does it need to travel to pass through without being hit?

 

[BvU]

And 22 cm is 0.22 m!
However: think if you need it

 

[Ursa]

To be perfectly frank, your approach in the OP just looks weird and confused to me. For example, it is unclear to me what your 35.7 rad is. It is certainly not the rotation needed for the wheel to turn between spokes (which must be an eighth of 2pi, which is pi/4 < 1 rad).This is correct. So, given that the arrow is 26 cm long, how fast does it need to travel to pass through without being hit?

$\Delta x = vt$ (for a=0) so $0.26 m/ 0.0357 s = 7.28 m/s$

 

[Orodruin]

$\Delta x = vt$ (for a=0) so $0.26 m/ 0.0357 s = 7.28 m/s$

Consider the number of significant digits in your result. Otherwise yes.

 

[rude man]

Depends on the position of the wheel when the arrow hits the plane of the spokes.

If it enters just before the next spoke, the answer is infinity!

(Just in a silly mood which some people may not appreciate, judging from past history.)