Question #257525

The average angular acceleration of a s tone lodged in a rotating car wheel is 100 rads -2 what will the stones final angular speed be after 3.0 s if it starts from rest ? Also calculate it's angular displacement during this time

Expert's answer

Explanations& Calculations


  • Assuming the stone would remain in contact with the tyre without dislodging from it,

a)

  • Apply ωf=ωi+αt\small \omega_f = \omega_i +\alpha t for the motion, you can compute the final angular velocity.

ωf=0+100rads2×3.0s=300rads1\qquad\qquad \begin{aligned} \small\omega_f&=\small 0+100\,rads^{-2}\times3.0\,s\\ &=\small \bold{300\,rads^{-1}} \end{aligned}

b)

  • Apply θ=ωit+12αt2\small \theta= \omega_it+\frac{1}{2}\alpha t^2 for the motion, you can compute the angular displacement of the stone.

θ=0×3.0s+0.5×(100rads2)×(3.0s)2=450rad\qquad\qquad \begin{aligned} \small \theta&=\small 0\times3.0\,s+0.5\times(100\,rads^{-2})\times(3.0\,s)^2\\ &=\small \bold{450\,rad} \end{aligned}

  • This means the stone has completed a full rotation (2πrad=3600\small 2\pi \,rad = 360^0 ) and another quarter (π/2rad=900\small \pi/2\,rad =90^0) during this 3 seconds.

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