http://serc.carleton.edu/dmvideos/players/87028.html?hide_banner=true We are NOT
ID: 1517972 • Letter: H
Question
http://serc.carleton.edu/dmvideos/players/87028.html?hide_banner=true We are NOT using data from the video. The engine is mounted to the wheel at an angle = 20 degrees from a line tangent to the edge of the wheel as shown. You find that when you light the rocket, the wheel will spin a total of 3.75 revolutions in a time of 2.0 seconds while the rocket is still burning. What is the force (thrust) of the rocket, assuming the thrust is constant while it is burning? The spinning wheel has a mass of 2.9 kg and a radius of 0.35 m, is a uniform disk, and the mass of the rocket can be neglected. Give your answer in Newtons.
Explanation / Answer
from equation of rotatory motion
angular displacement = wo*t + 0.5*alpha*t^2
angular displacment = 3.75 rev = 3.75*2*pi rad
3.75*2*pi = 0 + 0.5*alpha*2^2
angular acceleration alpha = 11.8 rad/s^2
torque acting = r*F*sintheta
but torque = I*alpha
I = inertia = (1/2)*m*r^2
r*F*sintheta = (1/2)*m*r^2*alpha
0.35*F*sin20 = (1/2)*2.9*0.35^2*11.8
F = 17.51 N <<<<<<---------answer
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