Solving angular acceleration Problem

[pennywise1234]

Homework Statement

The absolute angle of the thigh has the following angular velocities during the support phase of
walking. Calculate the angular acceleration at time 0.02s in rad/s and in deg/s

Time (s) Angular Velocity (rad/s)
0 s ---- 1.033
0.02s ------ 1.511
0.04s ----- 1.882
0.06s -----2.19

Homework Equations

i used a= change in v/t but did not get the correct answer

The Attempt at a Solution

i subtracted the final velocity from the velocity at 0.02 and did the same with time. so i got 0.679/0.04 but my answer is off. what am i doing wrong ?

 

[Mastermind01]

i subtracted the final velocity from the velocity at 0.02 and did the same with time. so i got 0.679/0.04 but my answer is off. what am i doing wrong ?

By doing what you did you get the average acceleration over the time interval t = 0.02 to t = 0.06. You need the angular acceleration at t = 0.02 . I think your best bet is to subtract v at t = 0.04 from the initial velocity to get the average acceleration over the interval t = 0 to t = 0.04 which would be approximately equal to the acceleration at t = 0.02. Does that help?

 

[pennywise1234]

yes, thank you. by why minus from 0.04s- 0 and not from 0.06 - 0.02 ?

 

[Mastermind01]

The acceleration is not constant in this case which can be verified by doing a = v/t for different intervals. Therefore in this case a = v/t gives the average acceleration over the time interval t. So for the acceleration at t = 0.02 we need the average acceleration for an interval which include t = 0.02. The best estimate would be the interval t = 0 to t = 0.04. The interval t = 0.02 to t = 0.06 may contain 0.02 but it's at an extreme thus the average over that interval does not represent the acceleration at that point.

Hope this helps.

 

[pennywise1234]

thank you. i wanted to ask you when dealing with constant acceleration what equation do you use? do you use

v(final velocity)= u(initial velocity) + at (acceleration multiplied by time)

 

[Mastermind01]

Yes. The equation is the same as the one we use except ours will give average acceleration as the acceleration is variable.