Angular Displacement at Specific Times

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mattmannmf
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a) What is the angular displacement of the wheel at t = 0.7 sec with angular velocity of -6.9?

b) What is the angular displacement of the wheel at t= 3.5 sec with angular velocity of 1.5?

I thought all i did was to multiply (.7)(-6.9) to get angular displacement for the first one, that's wrong. so not sure anymore.

angular velocity= angular displacement/ time (right?)
 
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θ is usually expressed in radians / sec. So θ = ω * t looks ok unless the answer is looking for revolutions or degrees?
 
rad.. but how they wrote it , they gave me a graph with a line. y-axis was angular velocity, x was time. i figured out the equation y=3x-9. so at t=.7 that's angular velocity of -6.9, and t=3.5, velocity 1.5.. but there looking for rad.

but regarding a, i get -4.83, but that's wrong (.7* -6.9)
 
mattmannmf said:
rad.. but how they wrote it , they gave me a graph with a line. y-axis was angular velocity, x was time. i figured out the equation y=3x-9. so at t=.7 that's angular velocity of -6.9, and t=3.5, velocity 1.5.. but there looking for rad.

but regarding a, i get -4.83, but that's wrong (.7* -6.9)

They gave you a graph? And it has a slope? Then you don't have constant ω. You should have said so initially.

So that means then that you have to use the rotational analog to velocity equation for ω with your final ω = -2.1 by your statement ...

ωf2 = ωi2 + 2*a*θ

2.12 = 32 + 2*3*θ

See:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin
 
i am a little confused...where did the 2.1 come from?
 
mattmannmf said:
i am a little confused...where did the 2.1 come from?

Sorry. Apparently so am I. I scribbled out my graph wrong from your description.

That should be initial ω of -9 and final ω of - (9 - 2.1) or -6.9

The method is the same though.
 
AHHHhhh, ok! makes sense... initial velocity is when x=0..haha ok