Rotational Kinematics of bicycle breaks

[tem_osu]

[SOLVED] Rotational Kinematics

Homework Statement

A person is riding a bicycle, and its wheels have an angular velocity of +18.0 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is +13.5 revolutions.

(a) How much time does it take for the bike to come to rest?

(b) What is the angular acceleration of each wheel?

Homework Equations

(a) θ = (ω0 + ω)t

(b) ω = ω0 + αt

The Attempt at a Solution

(a) +13.5 rev = (18.0 rad/s - 0 rad/s)t
t = (+13.5 rev)/(18.0 rad/s) = 0.75 - wrong answer

(b) (ω - ω0)/t = α
(0 rad/s - 18.0 rad/s)/0.75 = α - wrong answer


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Homework Statement

After 10.0 s, a spinning roulette wheel at a casino has slowed down to an angular velocity of +1.70 rad/s. During this time, the wheel has an angular acceleration of -5.05 rad/s^2. Determine the angular displacement of the wheel.

Homework Equations

ω^2 = ω0^2 + 2α(θ − θ0)

The Attempt at a Solution

(+1.70 rad/s)^2 = 0 + 2(-5.05 rad/s^2)θ
θ = [(1.70 rad/s)^2]/2(-5.05 rad/s^2) = -0.286 - wrong answer

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Please help me, I don't know what I'm doing wrong.

 

[Dick]

Start with 1). Your equation a) is wrong. This is accelerated motion. The equation should be:
theta=omega_0*t+(1/2)*alpha*t^2. Solve your equation b) for alpha and substitute that into a). Now the only variable is t.

 

[slider142]

Homework Statement

A person is riding a bicycle, and its wheels have an angular velocity of +18.0 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is +13.5 revolutions.

(a) How much time does it take for the bike to come to rest?

(b) What is the angular acceleration of each wheel?

Homework Equations

(a) θ = (ω0 + ω)t

(b) ω = ω0 + αt

The Attempt at a Solution

(a) +13.5 rev = (18.0 rad/s - 0 rad/s)t
t = (+13.5 rev)/(18.0 rad/s) = 0.75 - wrong answer

The equation you have listed as (a) is only valid for uniform angular velocity. Since the wheel is slowing down, velocity is not constant, so you have to use an equation for uniform acceleration (since the problem states that the wheel comes to a uniform stop).Since the acceleration is uniform, you can assume that \alpha = \frac{\Delta\omega}{\Delta t}.

 

[tem_osu]

equation 1: θ = ω0t + 1/2αt^2

equation 2: ω = ω0 + αt
(ω - ω0)/t = α

solve for α from equation 2 & plug into equation 1:
θ = ω0t + 1/2[(ω - ω0)/t]t^2

θ = ω0t + 1/2[(ω - ω0)]t
13.5 = 18.0t + 1/2(0 - 18.0)t
13.5 = 18.0t - 9t
13.5 = 9.0t
t = 1.5 seconds - but it's still wrong

Did I simplified it wrong?

 

[Dick]

A 'rev' is not a 'rad'. How many radians is 13.5 revolutions? Other than that, well done.

 

[tem_osu]

(a) 13.5 x 2Pi = 9.0t
t = 9.42 s

(b) (0-18.0 rad/s)/9.42s = -1.91 rad/s^2

Thank you so much.

I'm still confuse about the second problem though. I tried several ways, but I've gotten some weird answers.

 

[Dick]

omega_0 in that problem isn't zero. The wheel starts at velocity omega_0, slows down at a rate of -5.05rad/sec^2 for 10 sec. Then it's speed is 1.7rad/sec. Can you figure out the correct value for omega_0. Think about it...

 

[tem_osu]

ω = ω0 + αt
ω - αt = ω0
(1.70 rad/s) - (-5.05 rad/s^2)(10.0 s) = ω0
ω0 = 52.5 rad/s

ω^2 = ω0^2 + 2αθ
(1.70^2) = (52.2^2) + 2(-5.05)θ
θ = [(1.70^2) - (52.2^2)]/(2 x -5.05) = 269.5 rad

GOT IT!
Thank you so much for helping me. ^-^