How Do You Calculate the Final Angular Speed of a Bicycle Wheel?

[nightshade123]

[SOLVED] Angular Speed

\tau = r * F

Homework Statement

youve got our bicycle upside-down for repairs with its 66 cm diameter wheel spinning freely at 230 rpm. the mass of the wheel is 1.9 kg and is concentrated mostly at the rim. you hold a
wrench against the tire for 3.1s with a normal force of 2.7 N. if the coefficient of friction between the wrench and the tire is 0.46 what is the FINAL ANGULAR SPEED OF THE WHEEL?

Homework Equations

diameter = 66 cm
radius = .033m
mass = 1.9 kg
Norm Force = 2.7 N
friction coef = .46
\omega0 = 230 rpm
\omega0 = 24.1 rad/s
w = ?

eqns
I = Mr^2

\omega = \omega0 + \alpha * t

\tau = r * F * sin(theta)

\tau = I \alphav=\omega * r

v=\omega * r

The Attempt at a Solution

x

\tau = r * F

where F is the Friction Force

r*F = M * r^2 * \alpha

we can solve for \alpha and it = 19.08

\omega = ( \omega0 + \alpha * t )\omega = 85.48 rad/s
that means it sped up...

idk how to do this problem keep getting confused with the answers I am getting

any help?

thanks in advance for your time and effort.

 

[physixguru]

initial angular speed= 230 rpm.

torque = r * f * sin 90 degrees

>> torque= .33 * .46 * 2.7 ( radius * frictional force)

calculate torque.

Now the big ques is:
What is the moment of inertia of a wheel?

 

[Doc Al]

Homework Equations

diameter = 66 cm
radius = .033m

You have a typo here that messed you up later. Radius = 0.33m.

The Attempt at a Solution

x

\tau = r * F

where F is the Friction Force

r*F = M * r^2 * \alpha

we can solve for \alpha and it = 19.08

Perfectly correct, except that you used the wrong value for the radius.

Note that the angular acceleration created by the friction is opposite to the initial angular velocity, so the wheel slows down.

 

[nightshade123]

im having trouble interpreting my answer

\omega0 = 24.1 rad/s

asuming that is right

\omega = \omega0 + \alpha * t

then i put it in this eqn to find \omega = 30.238 rad/s

so next i would take \omega - \omega0 to find the change over 3.1 sec?

thus producing an answer of 6.138 rad/s

and then they want to know final angular speed so 24.1 - 6.138 = 17.962 rad/ s^^^^that was the more logical approach ^^^^vvvv you can also say that vvvv

this one just produces the same answer negative, so that's why it doesn't make sense

\omega = 24.1 rad/s

\omega = \omega0 + \alpha * t

solve for \omega0 and you get -17.962 rad/s, but then you got to change the sign, which u can because it asks for speed

 

[Doc Al]

im having trouble interpreting my answer

\omega0 = 24.1 rad/s

asuming that is right

That's the initial angular speed.

\omega = \omega0 + \alpha * t

then i put it in this eqn to find \omega = 30.238 rad/s

As I already pointed out, the acceleration is negative. Done correctly, this will give you the final speed.

so next i would take \omega - \omega0 to find the change over 3.1 sec?

thus producing an answer of 6.138 rad/s

The change is just alpha*time. What is your value for alpha?

and then they want to know final angular speed so 24.1 - 6.138 = 17.962 rad/ s

Seems like you're doing a bit of extra work.

vvvv you can also say that vvvv

this one just produces the same answer negative, so that's why it doesn't make sense

\omega = 24.1 rad/s

\omega = \omega0 + \alpha * t

solve for \omega0 and you get -17.962 rad/s, but then you got to change the sign, which u can because it asks for speed

Not sure what you're doing here. Why would you solve for the initial speed? That's given.

 

[Google_Spider]

im having trouble interpreting my answer

\omega0 = 24.1 rad/s

asuming that is right

\omega = \omega0 + \alpha * t

then i put it in this eqn to find \omega = 30.238 rad/s

\omega will not increase. It will decrease. You have to put a negative sign before \alpha in that equation.

 

[nightshade123]

that makes sense! and it produces the 17.962 rad/s instantly! thanks for the help

doc al, since i forgot to change accel to negative i showed how to produce the same answer, but with a lot more work, lol and i was just throwing out that idea for the 2nd part, i knew it didnt make sense but it was worth mentioning.