How Do You Convert Angular Distance from Radians to Linear Distance?

[Sparky500]

Homework Statement

A wheel starts from rest and accelerates uniformly with an angular acceleration of 4rad/〖sec〗^2. What will be its angular velocity after 4 seconds and the total distance traveled in that time?

Homework Equations

The Attempt at a Solution

i have got my answer and the unit is rad, i know that rad can be converted into degrees or even revs, but what about units in lengths (m)? or is the unit rad correct in my answer

thanks

 

[malawi_glenn]

You must know the radius of the weel.

 

[Sparky500]

thanks, so if i have not got the radius of the wheel is my answer of 40rad correct?

 

[malawi_glenn]

thanks, so if i have not got the radius of the wheel is my answer of 40rad correct?

How did you get 40rad ?

 

[Sparky500]

this is how i calculated the above:

ω=0+4 rad/〖sec〗^2 x 4sec

ω=0+16 rad/sec

ω=16 rad/sec

θ=1/2 (w_1+w_2 )t

=1/2 (20 rad/sec)t

=10 rad/sec x 4sec

=40rad

 

[malawi_glenn]

\omega = \omega _{0} + \alpha t

\theta = \theta _{0} + \omega _{0} t + \frac{1}{2} \alpha t^{2}

Now if alpha is constant 4
and time t is 4, what do we get?

 

[Sparky500]

ω=16 rad/sec (as shown in my answer)

the 40rad is my total distance (does this need to be converted)?

 

[malawi_glenn]

yes but the total angle covered is 32 rad

theta = 0 + 0 + 0.5 * 4 * 4^2

 

[Sparky500]

can you expain that a bit more as i don't know what you mean (sorry) so 40 rad is not the distance?

 

[malawi_glenn]

\theta = \theta _{0} + \omega _{0} t + \frac{1}{2} \alpha t^{2}

This is the formula, right?

The given values is alpha = 4 and time = 4. The initial angular veolcity is 0(starts from rest) and the theta_zero is also 0...

 

[Sparky500]

\omega = \omega_{1} + \alpha t

this is the formula given in class (if it can be read)

 

[malawi_glenn]

yes that gives you the new angular velocity, but you was also asking of the "lenght" covered by the weel in that time (4s)...

The total "lenght", is given by the forumula i gave you in post #10, where omega_zero is the initial angular velocity (of course). Now what is the anser for theta?

 

[Sparky500]

i am going to start this one over again using the information you have provided and repost what i get. Coul di ask you to have a look at this post for me https://www.physicsforums.com/showthread.php?t=174374 and let me have your thoughts?

thanks

 

[Sparky500]

i have now worked out that angular velocity = 16rad/sec

using \theta = {1/2} (\omega_{1} + omega_{2})

i calculated the distance to be 32rad

please confirm.

 

[malawi_glenn]

yes that it correct, see my post #8 otherwise...

 

[Sparky500]

thanks you have been a great help :)

 

[malawi_glenn]

good. this is a good site, and also wikipedia can be good sometimes too

http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html