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I have another physics problem and I'm trying to solve this using formulas I found on the web. I'm an online student and there is no textbook given for our course nor is the topic explained in the lessons so I'm trying to educate myself watching youtube videos and reading stuff online etc...

A fortune wheel is 2.6 meters in diameter (r = 1.3m). The initial velocity is 2m/s. After rotating 540 degrees, the velocity is 0 (wheel stops). What is the angular acceleration of the wheel?

What I learned online is that the angular velocity is calculated as v/r = 2/1.3 = 1.54rad/s When I know the angular velocity, I can solve for the time since one of the formulas I found is w = theta/t so t = theta/w. Once I have the time I can solve for the angular acceleration: a = w/t. Is this correct so far? If yes, I have a question concerning the theta value. Is this really in degrees or rads? If the theta value is degrees, the time would be 540/1.54 = 350.65s. This number seems wrong so I wonder if there are any mistakes in the formulas or if the degrees have to be converted into rads?

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Ordinarily, physics in general is done using SI units, but that's not absolutely necessary. It's just that the equations start needing odd correction factors. So, yes, convert degrees into rads to keep things simple.

Your problem is badly stated: "The initial velocity is 2 m/sec." almost certainly means that the tangential velocity is 2 m/sec, since that is the most obvious interpretation for this sort of problem. But there's no iron-clad guarantee, since it "might" mean that the tangential velocity at, let's say, a radius of 0.5 meters (1 m diameter) is what's being established. Granted, this would be unduly confusing, but since the radius corresponding to your velocity is not stated, it's impossible to be absolutely certain what is intended. But let's go with tangential.

So you can compute initial angular velocity $\omega$, and you know the final angle $\theta$ (converted to radians, of course). Just as linear motion is described by $$s = \frac{at^2}{2}$$ so $$\theta = \frac{\alpha t^2}{2}=\frac{\omega t}{2}$$ since for constant acceleration $\omega = \alpha t$, and $$t =\frac{2\theta}{\omega}$$Once you find t, you can use that to find $\alpha$

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  • $\begingroup$ This confuses me... Why is t equal to 2*theta/w? All the information I find on angular velocity says that w = theta/t which means t = theta/w and not 2*theta/w. Btw. the initial velocity comes from the "contestant spinning the wheel" in my assignment. So my calculated angular acceleration is .252rad/s^2 since t = 9.43(theta)/1.54(w). Am I right? $\endgroup$ – user102280 Dec 27 '15 at 17:48
  • $\begingroup$ @mhenkes92 = Look at my second equation. Rearrange the first and third terms. Theta = omega t /2. Do you see where the 2 comes from? Do you see why it cannot just disappear? $\endgroup$ – WhatRoughBeast Dec 27 '15 at 18:29
  • $\begingroup$ yes, I see where it comes from. It's just weird to me that this page states w = theta/t: formulas.tutorvista.com/physics/angular-velocity-formula.html So which is wrong? $\endgroup$ – user102280 Dec 27 '15 at 18:44
  • $\begingroup$ You are confusing two uses of theta. Notice that for a constant angular velocity, $\theta = \omega t$ (just as travelling for one hour at 60 mph moves you 60 miles). However, in this case theta is not constant, so the relationship does not hold. $\endgroup$ – WhatRoughBeast Dec 27 '15 at 18:49
  • $\begingroup$ ok, I see. But that doesn't change anything about the equation that w = v/r , does it? $\endgroup$ – user102280 Dec 27 '15 at 19:01
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The angular velocity, in the case of uniform acceleration is given by:

$$\omega(t)=\omega_0+\alpha t$$

And:

$$\theta(t)=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2$$

Also:

$$v=\omega R$$

Here, calculate $\omega_0$ from the initial tangential speed and radius, then set $\omega(t)=0$ and $\theta_0=0$, so we have:

$$0=\omega_0 + \alpha t$$

$$\alpha=-\frac{\omega_0}{t}$$

We also know that $540\:\mathrm{degrees}=3\pi$

$$3\pi=\omega_0 t-\frac{1}{2}\frac{\omega_0}{t}t^2$$

$$3\pi=\frac{1}{2}\omega_0 t$$

$$t=\frac{6\pi}{\omega_0}$$

Thus:

$$\large{\alpha=-\frac{\omega_0^2}{6\pi}}$$

$\alpha <0$ because it's a deceleration.

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