Angular velocity at 1/2 radius

[pawlist]

Homework Statement

angular velocity about the axle = 65 rads^-1
diameter = 0.54m
radius= 0.27m
1/2radius= 0.135m[/B]

Homework Equations

v=rw
w=θ/t
a=v^2/r=w^2r
(Might be the probability that i have to use another equation)[/B]

The Attempt at a Solution

Firstly i cut the diameter in half which gave me the radius of 0.27m
Then i cut it in half again(since i need ang.velocity between axle and outer edge)
which gave the radius of 0.135m
I am assuming that v=constant on all points on the wheel(this is where i may be wrong)
so what i did was v=0.27m*65rads-1
which gives v=17.55ms-1
since i am assuming v=constant everywhere
i arranged the formula w=v/r
which i did 17.55/0.135m=130rads-1
anyone can help me because I am almost sure that i have a wrong answer

Thanks[/B]

 

[MalachiK]

If the wheel is going to stay in one piece, shouldn't all points along the radius complete one revolution in the same amount of time?

You've assumed that the linear speed v is constant all along the radius, but are you sure that this is correct?

 

[pawlist]

I don't know if its correct since i don't have an answer sheet

 

[MalachiK]

Sure, but do you agree that the angular speed must be the same at all points along the radius?

 

[pawlist]

So what you are saying is that w is 65 rads-1 at all points?because thr question is asking me to find w at half the radius

 

[MalachiK]

I believe that is correct. The thing to understand is that ω =2π/T, or in other words T = 2π/ω. For the wheel to stay in one piece, all of the points along the radius must complete each revolution in the same amount of time. No matter where you are along the radius, the time for one rotation must be the same. If the edge of the wheel completed one revolution in less time than the point at half the radius then the wheel would break apart! A higher angular speed at the outer edge would mean that the rim of the wheel was turning through a larger angle each second the bits of the wheel that are closer to the centre.

In order for this to happen, the linear speed v must vary along the radius. The further away from the centre you are the faster you must travel in order to complete one revolution in the same amount of time as the points that are closer to the centre.

 

[pawlist]

But doesn't the outer part of the wheel cover more diatance than the inner part.It is true that they both take T to complete 1 revolution but the outer part is covering more diatance

 

[MalachiK]

That's correct. The distance traveled by a point on the outer edge of the wheel during one revolution is larger. The point must move through more metres per second to complete one revolution in the same time as a point closer to the middle. The linear speed v is larger on the edge of the wheel (more metres each second) but the angular speed ω must be the same (it turns through the same angle each second as the points closer to the centre).

 

[pawlist]

So basically
w-constant
v-variable?

 

[MalachiK]

Yep. In fact you have already written this down in your original post... v = ωr. Since the wheel is rigid, ω must be the same everywhere. On the other hand, v gets larger as you move out from the centre (v is proportional to r).

 

[pawlist]

So now i moved on to the second question and i did
v=wr
v=65*0.27
V=17.55ms-1
But it needs to be km/hr and the answer must be around 63
However
m to km *1000
so 1000*17.55=17550
To get seconds to hrs you do /3600
So 17550/3600=4.875
What ami doing wrong?

 

[MalachiK]

You've got the correct speed in metres per second, but your conversion into kilometres per hour is wrong.

If you're moving at 17.55 m s^-1, how far will you move in one hour?

 

[pawlist]

Ohhhhhh
3600*17.55=631800/1000=63.18km/h-1 am i correct?

 

[MalachiK]

That's what I get on my calculator, and it rounds to 63 km hr^-1 as required by the question. I think it's safe to say that you've got the correct answer.

 

[SteamKing]

And a tip about posting images for problem statements: Please take the picture in plenty of light so that others can read the image. The image in the OP looks like it was taken under a blanket at night in a closet.

 

[pawlist]

I think its because i had my fladh on

 

[pawlist]

And in another section it says that the motorcycle crosses a bridge just without losing contant with the ground
Calculate the radius of curvature of the bridge

 

[MalachiK]

I take it that the motorcycle follows a circular path as it goes over the bridge? There's got to be some centripetal resultant force acting for this to happen.

Try drawing a diagram showing the motorcycle at the highest point on the bridge and then draw in all the forces.

 

[pawlist]

I used the equation r=v^2/g
17.55^2/9.81=31.4m
the answer seems bit big for an arc of a bridge what do you think?

 

[pawlist]

Its the same question no mass or forces are given just the diameter ,angular velocity and the velocity we have just found.
And g=9.81ms-2 presumably