Finding centripetal acceleration

[pinkerpikachu]

1.The blade of a windshield wiper moves through an angle of 90 degrees in .28 . The tip of the blade moves on the arc of a circle that has a radius .76m. What is the magnitude of the centripetal acceleration at the tip of the blade.



2. Ac= v^2/r



3. okay, the acceleration should equal r(theta)/t

but whenever I plug in my variables and solve I get an answer in the 7000s which is not reasonable at all.

The actual answer is 24 m/s^2 but I have no idea how to get there.

 

[Galileo's Ghost]

3. okay, the acceleration should equal r(theta)/t

This is not correct. This expression would give you the linear velocity at the end of the wiper blade.

 

[pinkerpikachu]

really? that is an equation my physics teacher and i derived together.

V = Δs/Δt = (rΔ(theta)/Δt)^2

then, plugging that into Ac

v^2/r = (rΔ(theta)/Δt)^2 / r = rΔ(theta)/Δt

 

[Mentallic]

Using $$A_{c}=\frac{v^{2}}{r}$$ It is clear you need to find the velocity since you already are given the radius.

The blade of a windshield wiper moves through an angle of 90 degrees in .28 .

I will assume that it is 0.28 seconds. The question states that it moves through an angle of 90^o. Therefore, the distance the tip of the blade moves is 90^o through a circle.

Using the circumference for a circle formula: $$C=2r\pi$$ where C=circumference, r=radius; This is the distance traveled around an entire circle, but we only want the distance covered in 0.25 of the circle.

Therefore, $$\frac{C}{4}=\frac{1}{2}\pi(0.76)$$

The distance covered for the tip of the wiper is $$d=0.38\pi . m$$

Using $$v=\frac{d}{t}$$ where v=velocity, d=distance (not displacement), t=time

Plugging in distance and time - $$v=\frac{0.38\pi}{0.28}$$

$$v=\frac{19}{14}\pi . ms^{-1}$$

Plugging back into the centripetal acceleration formula - $$A_{c}=\frac{(\frac{19}{14}\pi)^{2}}{0.76}$$

Hence, $$A_{c}\approx24ms^{-2}$$

 

[pinkerpikachu]

Thanks! It seems simple once I get to look at all the work.

Now I just really wonder what my physics teacher was going on about...the equation certainly didn't get me the right answer.

 

[Galileo's Ghost]

v = 0.38pi/.28
...and this is the linear velovity, r(theta)/t, not the acceleration.


Here is where you are making an error...

v^2/r = (rΔ(theta)/Δt)^2 / r = rΔ(theta)/Δt

you did not square the theta or the t

 

[Mentallic]

Well if you are looking for an equation that can find the centripetal acceleration straight from this kind of information given, then: (oh I will be skipping a few of the simpler steps since you seem more intelligent than I first suspected )

$$C=2r\pi$$

Hence, $$d=\frac{C\Theta}{360}=\frac{r\pi\Theta}{180}$$

Therefore, $$v=\frac{r\pi\Theta}{180t}$$

Therefore, $$A_{c}=\frac{(\frac{r\pi\Theta}{180t})^{2}}{r}$$

Finally, the equation you were looking for to plug and chug instantly is:

$$A_{c}=\frac{r(\pi\Theta)^{2}}{(180t)^{2}}$$

where,
$$r =radius (metres)$$
$$\Theta =Angle of turn (degrees)$$
$$t =time (seconds)$$