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AH
Carnegie Mellon University

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Problem 8

A wheel is rotating about an axis that is in the $z$-direction.The angular velocity $\omega_z$ is $-$6.00 rad/s at $t =$ 0, increases linearly with time, and is $+$4.00 rad/s at $t =$ 7.00 s. We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at $t =$ 7.00 s?

(a) The angular acceleration is positive
(b) $t=-\frac{\omega_{0 z}}{\alpha_{-}}=-(-6.00 \mathrm{rad} / \mathrm{s}) /\left(1.429 \mathrm{rad} / \mathrm{s}^{2}\right)=4.20$
(c) $\frac{-6.00 \mathrm{rad} / \mathrm{s}+4.00 \mathrm{rad} / \mathrm{s}}{2}=-1.00 \mathrm{rad} / \mathrm{s} . \quad \theta-\theta_{0}=\omega_{\mathrm{av}-z} t$

## Discussion

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## Video Transcript

So the question is asking us to find the sign of angular acceleration. So and then we're trying to find also win the angular velocity changes direction and as well as the angular displacement throughout the entire seven seconds. So first write are given. So we have the angular velocity initially was going to be negative. Six radiance per seconds. That t seconds Then we have final. So this means that for a the angular acceleration is going to be Yeah, the derivative of the angular velocity with respect to time. And we know that angular velocity is increasing linearly with respect to time, So this will be positive. So that means that the angular acceleration is positive throughout the entire seven seconds and then for B, they're asking us when does this change direction? So we should find the angular acceleration by taking out ah, Omega over adult ity. This is going to be equal to four minus negative six all over seven. And at this point, we can set up on equation and say that angular velocity he's going to be equal, Tio the the anger of acceleration rather is going to be all to the angular velocity final minus angular velocity initial all over, um T final minus t initial weariness. A teen initial zero and we're trying to find when the angular velocity trainers direction so will say that the final angular velocity is also zero and we're left with negative omega. It's the initial over T final, and then we can simply solve. So t Final would equal negatives Angular, glossy initial all over. How so, Xena? And that's going to be equal to six. Ten over seven, four point two seconds. So what this really means is that Omega Z is going to be less than zero from zero seconds, brother two four point two seconds and we're not, including four point two seconds, because the angular velocity at four point two seconds is going to be zero So and then we have angular velocity is gonna be creator than zero from four point two seconds two seven seconds. And we're including seven seconds and again, we don't include four point two seconds. So this would be your answer for Part B, and then let's get a new workbook. And if they're part see, we're asking for the total angular displacement so angular his basement can be derived from the formula. Ah, displacement final equals displacement initial plus thie initial angular velocity times time and plus half times the angular acceleration times times squared. And then, at this point, we Khun, simply substitute in for our known values. The initial angular displacement is going to be considered zero so we can say negative six times seven plus one over two. And we'LL keep ten over seven and I will have seven squared. And this is unequal native six point nine seven radiance. So this will be your answer for sea. And it's simply from the from the first from the first equation for angular for the final angular displacement. And that's the end of the solution. Thank you for watching.