Angular Motion Question Understanding help ?

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The discussion revolves around calculating the angular velocity and acceleration of a figure skater completing a double axle in 0.5 seconds. The calculated angular velocity is 1440 deg/s or approximately 23.132 rad/s, while the average angular velocity is debated, with some stating it is 0 due to the skater stopping. The angular acceleration is calculated as 960 rad/s², but confusion arises regarding whether it should be positive or negative, with clarification that it should be negative since the skater decelerates to a stop. The distinction between average angular acceleration and angular acceleration is also discussed, emphasizing that average values apply when motion is not uniform. Overall, the thread seeks clarity on the calculations and the correct interpretation of angular motion concepts.
uaeXuae
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Homework Statement



A figure skater completes a double axle (2 complete rotations) in 0.5 seconds. Calculate
the skater’s angular velocity and average angular velocity in a) deg/sec, and b) rad/sec.

If the skater manages to stop spinning in a time of 1.5 seconds,
what was the angular acceleration and average angular acceleration during this period (in deg/s)?

Homework Equations



Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)
Angular Velocity = theta/time

Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)
Angular Acceleration => Alfa = (w)/(t)
wf=wi + alfa*(t)

The Attempt at a Solution




Angular Velocity => w=(360*2)/0.5 = 1440 deg/s
1440 * (pi/180) = 23.132 rad/s

Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)
Average Angular Velocity => w=(360-360)/0.5 = 0 rad/s


Angular Acceleration => Alfa = (w)/(t)
Angular Acceleration => Alfa = 1440/1.5 = 960 rad/s^2

Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)= (0-1440)/1.5 = -960rad/s^2

wf=wi + alfa*(t) = > alfa = (wf-wi)/t ==> alfa = (0-1440)/1.5 = -960 rad/s^2


im almost sure about my results in the (Angular Velocity and Average Angular Velocity ) but for the (Angular Acceleration and Average Angular Acceleration i am not )

could someone correct my answers and explain the changes that has been made.

Thanx in Advance.
 
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One is correct. One revolution is 360° or 2\pi radians, and one correctly used the relationship \pi/180 rad/deg.
 
Thanx but can anyone explain what is the correct angular acceleration (+960 or -960)and why ? and what is the difference in the pronouncings Average angular acceleration and angular acceleration hence

average speed = total distance / time.
average velocity = displacement / time.
 
Last edited:
<up>
 
Average velocity or acceleration is calculated if they are not uniform. In your problem there is no indication of that.
 
rl.bhat said:
Average velocity or acceleration is calculated if they are not uniform. In your problem there is no indication of that.

What do u mean not uniform ? and what is the indication or the thing that will make me know wheather its uniform or not ?
http://img516.imageshack.us/img516/3300/1234rr6.jpg

i still don't know how the second part is solved. Can someone clarify things to me ?
 
Last edited by a moderator:
Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)This is not true.
average angular velocity = total angular displacement / total time.
 
ok
applying the law
average angular velocity = total angular displacement / total time.

average angular velocity = 1440/ 1.5 = 960


how is it -960 ?
 
rl.bhat said:
Average Angular Velocity => w=(Theta2-Theta1)/(t2-t1)This is not true.
average angular velocity = total angular displacement / total time.

and what would that


angular velocity be equal to in this case ?!
 
  • #10
Angular Velocity => w=(360*2)/0.5 = 1440 deg/s
1440 * (pi/180) = 23.132 rad/s. = Average angular velocity
Angular Acceleration => Alfa = (w)/(t)
Angular Acceleration => Alfa = 1440/1.5 = 960 rad/s^2

When the body comes to rest the acceleration cannot be positive.
 
  • #11
uaeXuae said:
ok
applying the law
average angular velocity = total angular displacement / total time.

average angular velocity = 1440/ 1.5 = 960


how is it -960 ?
As one worked out initially - Average Angular Acceleration => Alfa = (w2-w1)/(t2-t1)= (0-1440)/1.5 = -960rad/s^2

wf=wi + alfa*(t) = > alfa = (wf-wi)/t ==> alfa = (0-1440)/1.5 = -960 rad/s^2

The skater starts with an initial angular velocity wi at ti, and then decelerates to wf at tf.

The change in angular velocity is wf-wi and the change in time is tf-ti, and the angular acceleration is alfa = (wf-wi)/(tf-ti). If the body comes to rest, wf=0, to alfa = -wi/(tf-ti), and since tf > ti, the difference is positive, to the angular acceleration is negative.


w=(Theta2-Theta1)/(t2-t1) is correct, but one must be careful that Theta2 and Theta1 represent cumulative angular displacements from the same reference angle, and not just the angular displacement on a circle, i.e. Theta2 and Theta1 could > 360° or 2pi rad. In the given expression Theta2-Theta1 is the total angular displacement occurring between t2 and t1.
 
  • #12
thanx for your help but sorry for insisting ...

rl.bhat said:
Angular Velocity => w=(360*2)/0.5 = 1440 deg/s

Understood
rl.bhat said:
1440 * (pi/180) = 23.132 rad/s. = Average angular velocity

Thats just the same its converting from degree to radian( so converting from degree to radian is average angular velocity

rl.bhat said:
Angular Acceleration => Alfa = (w)/(t)
Angular Acceleration => Alfa = 1440/1.5 = 960 rad/s^2

Not sure about that.

rl.bhat said:
When the body comes to rest the acceleration cannot be positive.

Understood.


Once again why was the average angular velocity = 0 but there was a valus for the angular velocity. Not only that but what law should be used when calculating
Angular Acceleration & average Angular Acceleration.
 
Last edited:

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