Angular speed, acceleration, and angle of a Ferris wheel

[ac7597]

Homework Statement

The county fair features a big Ferris Wheel of radius R=8 m. The drive mechanism is designed to accelerate the wheel from rest to a maximum angular speed ω=1.3 radians per second in a gradual manner: the angular speed at any time t is given by:

ω(t)=(1.3)∗(1.0−e^(−t/τ) )
where t is the time in seconds since the ride started,
and τ=22 seconds is the so-called "time constant" of the ride; it indicates roughly the time it takes for the ride to change its speed significantly.

What is the angular speed ω of the Ferris Wheel at time t=10 seconds after it has started from rest?

What is the angular acceleration at time t=10 seconds?

What is the total angle by which the wheel rotates over this period of t=10 seconds?

How long does it take the wheel to complete its first revolution as it starts from rest?

Relevant Equations

ω(t)=(1.3)∗(1.0−e^(−t/τ) )

ω(10)=(1.3)∗(1.0−e^(−10/22) )= 0.475 rad/s

0.475 rad/s=0 +α(10second)
α=0.0475 rad/s^2

∫ω(t)=Θ =1.3t + 28.6e^(-t/22) | (t=10s, t=0)
total angle by which the wheel rotates over this period of t=10 seconds = 2.55 rad

Θ= 2(pi)(8m)= 1.3t + 28.6e^(-t/22)
0=1.3t + 28.6e^(-t/22) - 2(pi)(8m)
t=34 seconds

 

[ac7597]

angular acceleration is not 0.0475 rad/s^2. Also the time is not 34 second. I don't know why

 

[TSny]

angular acceleration is not 0.0475 rad/s^2.

In finding the angular acceleration, you used an equation for constant angular acceleration. Is the angular acceleration constant in this problem?

Also the time is not 34 second. I don't know why

Did you forget the contribution from the lower limit (t = 0) when integrating ω(t) to find θ?

 

[ac7597]

d(ω(t))=α(t)= 1.3e^(−t/22) /22
α(10)= 1.3e^(−10/22) /22=37.5E-3 rad/s^2

 

[ac7597]

Θ= 1.3t + 28.6e^(-t/22) |(t, t=0)
2(pi)= 2(pi)(8m)= 1.3t + 28.6e^(-t/22) -(1.3(0) + 28.6e^(-0/22))
2(pi)= 2(pi)(8m)= 1.3t + 28.6e^(-t/22) -(28.6)
0= 2(pi)(8m)= 1.3t + 28.6e^(-t/22) -(28.6)- 2(pi)
t=16.4 seconds