Acceleration of the cart on a Ferris Wheel (Circular Motion)

[Crunge]

Homework Statement

The tangential velocity ,vt, for the cart of a ferris wheel varies with time according to vt(t)= k*t. Assume that the constant k = 0,695 m/s^2 and that the radius of the ferris wheel is 11,7 m.

What is the sum of the cart acceleration after 3,32 seconds?

Relevant Equations

(The equations I have tried to use)
vt(t) = k*t
radial acceleration: (instantaneous) ar = (v^2) / t
angular velocity: ω = v/r
angular acceleration (average): α = Δω/Δt
angular acceleration (instantaneous): α = 1/T^2
T = 2π/ω
tangential acceleration: at = α*r

After 3,32 seconds, vt should have varied by 0,695*3,32. I have done a previous exercise where you only needed to calculate the radial acceleration in this scenario. There, I took the vt after the given time, squared it and then divided with the radius. I remember clearing that one, so in this exercise I would think to do the same. Take 0,695*3,32, square the product and divide by 11,7 (the radius). Ends up being roughly 0,455.

But then I am having trouble getting the right tangential acceleration. I have also done an exercise where you calculate tangential accelaration, which I completed, but in this one I can't seem to get the right answer no matter how I go about calculating the tangential acceleration.

First the formula for average angular acceleration: ω = v/r. In this problem, I think Δω should simply be (k*t)/r since the linear velocity varies according to k*t. Δt should be 3,32 seconds, according to what the problem defined. With this, I calculated α with Δω/Δt. If I then use at = α*r,
I as expected end up with 0,695, the same value as k. This seems suspicious, though it is undoubtebly where I end up using these equations. This is the result you get if you calculate the acceleration as linear as well, both average and instantaneous. (Δv/Δt), or d/dt * v(t). Though if I add the radial and tangential acceleration together I get the wrong answer.

I eventually tried to use the formula for instantaneous angular acceleration. ω is yet again (0,695*3,32)/11,7. T should then be 2π/ω, which I then square and inverse according to α = 1/T^2. Lastly I multiply with 11,7 according to at = α*r. This of course ends up with a suspiciously small number which also doesn't give the right answer when added to the radial acceleration.

From this point, try as I might, I can't think of another way to go about this problem. I feel like the strategies I tried first have worked before in other exercises. My best guess is that there's something about that formula vt = k*t, that I am not getting. Maybe the fact that it describes how vt varies has some significance that I am not picking up on.

 

[kuruman]

How did you add the tangential and centripetal accelerations? Don’t forget they are orthogonal components of the total acceleration vector.

 

[Lnewqban]

Welcome, @Crunge !
What is the official correct answer?

 

[Crunge]

Welcome, @Crunge !
What is the official correct answer?

−0,823 m/s² (rounded)

 

[Crunge]

How did you add the tangential and centripetal accelerations? Don’t forget they are orthogonal components of the total acceleration vector.

Oh I just added the results together, hopefully that's what I'm doing wrong

 

[topsquark]

Oh I just added the results together, hopefully that's what I'm doing wrong

Acceleration is a vector. So you need to add the centripetal acceleration vector to the tangential acceleration vector. Luckily, they are at right angles to each other. Do you recall how to do that?

-Dan

 

[Crunge]

Acceleration is a vector. So you need to add the centripetal acceleration vector to the tangential acceleration vector. Luckily, they are at right angles to each other. Do you recall how to do that?

-Dan

OHHHHHHHHHHH! Pythagoras that's right! Got the right answer now. Thank you so much! Feels kind of good that that's what I missed, at least understood what formulas to use.

 

[kuruman]

OHHHHHHHHHHH! Pythagoras that's right! Got the right answer now. Thank you so much! Feels kind of good that that's what I missed, at least understood what formulas to use.

The statement of the problem is misleading. It asks for "the sum of the cart acceleration". Unless this is faulty translation from a non-English language original, it should ask for the magnitude of the cart's acceleration. The use of the word "sum" is imprecise because it implies straightforward addition

Also, for future reference should this problem cross your path again, there is a simpler way to get the answer. The tangential component is
$$a_t=R~\alpha=R\frac{d\omega}{dt}=R\frac{ d(\frac{v}{R} ) }{dt}=\frac{dv}{dt}=k.$$You could have gotten this result by noting that only the tangential acceleration changes the speed in the tangential direction. For the centripetal component,
$$a_c=\frac{v^2}{R}=\frac{k^2t^2}{R}.$$Therefore, $$a=\sqrt{a_t^2+a_c^2}=\sqrt{k^2+\left(\frac{k^2t^2}{R}\right)^2}.$$

 

[LastScattered1090]

$$a_t= [...] =k.$$

Yeah! As I was reading, I was wondering why so many hoops were being jumped through to determine the tangential acceleration when it's basically given in the problem (v_t(t) is given to be a linear function)

 

[haruspex]

−0,823 m/s² (rounded)

Why negative? No reference direction has been specified, so shouldn’t it be a magnitude?

 

[kuruman]

Why negative? No reference direction has been specified, so shouldn’t it be a magnitude?

According to the statement of the problem we are looking for "the sum of the cart acceleration after 3,32 seconds". Looking for the magnitude was my assumption because it's the only quantity that makes sense and matches the given answer (without the negative sign). I think only the author of the solution can answer your question, or was it rhetorical?

 

[haruspex]

According to the statement of the problem we are looking for "the sum of the cart acceleration after 3,32 seconds". Looking for the magnitude was my assumption because it's the only quantity that makes sense and matches the given answer (without the negative sign). I think only the author of the solution can answer your question, or was it rhetorical?

Maybe the official answer is positive, but @Crunge ignored the sign.