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ConstantsPeriodic Table The angle througn which rotating wheel has turned in time t is given by sat_ b t2 c [4 where i5 in radians and { in seconds _...

Question

ConstantsPeriodic Table The angle througn which rotating wheel has turned in time t is given by sat_ b t2 c [4 where i5 in radians and { in seconds _

Constants Periodic Table The angle througn which rotating wheel has turned in time t is given by sat_ b t2 c [4 where i5 in radians and { in seconds _



Answers

A wheel rotates with an angular acceleration $\alpha_{z}$ given by $$ \alpha_{z}=4 a t^{3}-3 b t^{2} $$ where $t$ is the time and $a$ and $b$ are constants. If the wheel has an initial angular velocity $\omega_{0}$, write the equations for $(a)$ the angular velocity and $(b)$ the angle turned through as functions of time.

Mhm. This problem covers the concept of the can dramatic question for the traditional motion and this. To solve this problem first, we need to write the expression for the instantaneous velocity and this instantaneous acceleration. So by taking the time derivative, the instantaneous angular velocity, omega is for radiant four seconds plus six radiant All 2nd square into T plus three Radiance or 2nd Q into the square. That said, this is a question. And the expression for the analytic solution alpha is uh six gradients four, second square plus six radiant all second cube In two G. Let's say this is a question too. Now, for part a the angular velocity at Stephen equals from the equation one that we have derived for the uh and global ah city, you can substitute the value of Stephen into that equation. So by substituting four radian or second plus six radiant four seconds square Into T. and T. one equals two seconds less. Three radiant four second cube Into the square and there is two seconds square. So the travelocity two seconds comes out to me 28 radiance for second. No party. The anglo velocity at T. Two that is at four seconds equals from the same equation. We can write for radiance or second plus six radiance four seconds square into four seconds plus three radiance or 2nd square, sorry for a second. Q into four seconds square. So the angular velocity at the key to that is four seconds comes out to be uh scientist six radiance for a second. No party. The average acceleration there is the original acceleration is the changing uh angular velocity upon the changing tank. And between T. Two and T. Even the change in angular velocities 76 minus 28 radiance four seconds upon the changing time is full minus two seconds. So the average acceleration is uh 18. Sorry, every acceleration is 24, 24 radiance four seconds squared, no party the instantaneous acceleration alpha act even that is a two seconds equals from the second equation. We can write six radiant four seconds squared plus sixth radiant or second Q into no second. Are the angular instruments and good acceleration at even equals two secondaries 18 Radiance or 2nd square. Now part E. The angular acceleration alpha at T two equals six radiance four seconds squared plus six radiance four second cube Into that IMT two. And that is four seconds. So the instantaneous angular acceleration at T two equals 30 radiance for a second.

Mhm This problem covers the concept of the anglo velocity and they ignored acceleration. They little velocity because Omega Said, equals the time derivative of the angular displacement. Five or the angular velocity for part A equals D by detail 80 plus BT kills- C two due to the powerful are the angry velocity is a plus three beauties square -4 CTQ. No party. The anglo acceleration equals the time derivative of the and low velocity of the angular acceleration equals D by duty of a plus three. Bt square minus four City cute are the England accelerations six meaty minus 12 city square

This problem. It covers the concept of the rotational canna medics under in further acceleration. And to solve this problem for us where to find the initial and the velocity Uh at the beginning of the interval four seconds. So from the question we can write or McGovern equals theta minus half of al Fathi square upon T. Always substitute the value omega one equals Data is one. -1 of then. Re acceleration is three radiance for 2nd square. Into the time intervals. four seconds square upon four seconds Are the initial angular velocity at the beginning of the time interval is 24 radiance for a second. No, for the second part, We consider the final angular velocity at the beginning of the four seconds as 24 radiance or second. And for start, the initial angular velocity is zero radiance for a second. So the time spent before 4 2nd travel equals omega two minus omega one upon the constant angular acceleration, americans assured the value omega two is 24 radiance, or second minus the initial. That is zero radiance for second Upon the angular acceleration, and that is three radiance for a second, So the time spent before the 4 2nd travels comes out to be eight seconds.

This problem covers the concept of the charismatic property of generational motion. And to solve this problem fascinating right. The expression for the angle velocity and the acceleration and the angle velocity is the time they're ready of the anger displacement and by taking the time derivative, the expression for the velocity comes out today. Eight Radiance or 2nd square into T plus six radiant, four second Q. And to the square, Let's say this is a question too. And the expression for the angular acceleration comes out to be eight Radiance or 2nd square plus 12 Radiance for 2nd Q. And to G. Let's name the situation as equations now for part A at T one equals two radiant sorry, at even equals uh zero seconds several seconds from a question one. If we substitute the value of T one equals zero seconds. We get the England displacement is to radio. Now for part B. The angular velocity at jeevan from a question too. If we substitute the value of T0, we get zero radiance for second. Mhm. Now Barzee At T three equals four seconds from question to the angular velocity. That is omega T three equals eight Radiance or 2nd Square into four seconds plus six radiant 4 2nd q. And two four seconds square. Are the angular velocity at T3 equals 1 28 radiance for second. No party From the 3rd equation at T two equals uh two seconds. The angular acceleration that is Alfa T two equals eight Radiance or 2nd square plus 12 Radiance for second kill into two seconds. or the instantaneous acceleration at T two is um 32 radiance four seconds square. No party since the expression after and good acceleration depends upon the time. Therefore the angular acceleration alpha is not constant.


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