there is a useful grass and weed cutting tool that utilizes a 0.21-m length of nylon "string" rotating at 6200 rev/min about an axis perpendicular to one end of the string. (a) What is the time needed for the string to sweep out an angle of 35 degree? (b) Assuming that the length of the string does not change, find the distance through which the tip of the string moves during this interval.
6200 rev/min = frequency of rotation do you know formula for angular velocity ω ?
ω = θ / t
wright! but that is the definition of ω you will use it after you have calculate the ω now formula for ω involving frequency???
I don't know.
ω=2 π f
note you have to convert rev/min to rev/sec
Alright.
so we have ω=2*π* f ω=θ/t there for 2*π*f=θ/t t=θ/2πf f to rev/sec and θ to rad
Thanks let's see if I can solve it now.
you should find t=9.4086exp-4 sec
This is what I got for the first equation 6200 rev/min = 103.3 rev/s ω = 2π(103.3) ω=648.724
Am I correct ?
what you use for π ?
pi= 3.14
ok you are correct
how ever you could have a more acurate resalt if t=θ/2πf θ=(35/360)*2π rad f=6200/60 Hz t=....... t=(35*2π*60)/360*2π*6200 ...t=9.4086exp-4 sec this way you use your calculator only once at the end and have only one aproximation
in my place of the planet that is the way we do calculations don;t know if that is axeptable for you...
My teacher didn't teach me that way
ok folow you way This is what I got for the first equation 6200 rev/min = 103.3 rev/s ω = 2π(103.3) ω=648.724 rad/sec so far OK
now you have to convert degrees to rad do you know how?
35 degree = 0.6108652381965 rad
0.61055555555555555555rad
in your calculator: 35 / 360 = * 2 * 3.14 = 0.61055555555
now you have θ=0.610555 rad ω=648.724 rad/sec go for t
Alright thanks
I got 1062.5
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