angular speed=tangential speed/radius
solve for tangential speed.
b)assuming the acceleration was constant, then
angular accleration=(wf-wo)/time=
=(Vf-vo)/(r*time)
but time= distance(avg velocity)
=distance(wf+wo)/2 * radius
put that in for time, and you have it.
angular acc= (Vf-Vo)/(r*distance*(wf+wo)2radius)
angacceleration= 1/2*(vf^2-Vo^2)/r^2 * 1/distance
a bit more can come from this...
remember the formula
vf^2 =vo^2 + 2*a*d ?
That is a standard formula you should have memorized.
Its angular form is
Wf^2=Wo^2+2 *angacceleration*distance
Memorize that.
A dragster starts from rest and accelerates down the track. Each tire has a radius of 0.360 m and rolls without slipping. At a distance of 360 m, the angular speed of the wheels is 292 rad/s.
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(d) Determine the linear speed v of the dragster.
Number Unit
v = m/s
(e) What is the algebraic expression for the magnitude of the angular acceleration of the dragster's wheels? Express your answer in terms of the dragster's initial and final linear speeds, v0 and v, the distance x it travels, and the radius r of a wheel. (Answer using v_0 to be the initial linear speed, v to be the final linear speed, x to be the distance it travels, and and r to be the radius of the wheel).
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(f) Determine the magnitude of the angular acceleration of the dragster's wheels.
Number Unit
= rad/s^2
1 answer