**Description for Educational Website:**Two identical flywheels with a radius of 1.00 m are fixed in place on a table and spinning side-by-side, each spinning about their center. The setup is as follows:- **Wheel A:** - Initially spins in the clockwise direction. - Initial angular velocity (\(\omega_{A0}\)) = -16.0 rad/s. - Initial angular displacement (\(\theta_{A0}\)) = 0 rad. - Constant angular acceleration (\(\alpha\)) = 1.90 rad/s².- **Wheel B:** - Initially spins in the counter-clockwise direction. - Initial angular velocity (\(\omega_{B0}\)) = 16 rad/s. - Initial angular displacement (\(\theta_{B0}\)) = 0 rad. - Angular acceleration (\(\alpha\)) = -1.85 rad/s² - 5.33t rad/s³ + 1.02t² rad/s⁴.A diagram illustrates the wheels, with sensors indicating the direction of rotation. Both wheels have strings attached that presumably measure tension or angular displacement.**Task:**- **Graphs to Sketch:** a) **Angular Acceleration vs. Time:** - Plot the angular acceleration of both wheels over time. - Indicate the time at which their angular accelerations are equal. b) **Angular Displacement vs. Time:** - Plot the angular displacement of both wheels over the first ten seconds of motion.This exercise helps to visually compare the changes in angular motion over time for both wheels under defined mathematical conditions.

For a circular motion, the angular velocity is given by the relation ω=dθdt and angular acceleration is given by α=dωdt=ddtdθdt=d2θdt2 Therefore if α is known then θ can be easily found by integrating twice. In doing so, we will encounter two integration constants which can be determined using boundary conditions mentioned in the problem.For the first flywheel A we have initial displacement θA0=0 rad, the initial velocity ωA0=-16.0 rad/s and acceleration is constant given by αA=1.90 rad/s2. For the second flywheel B we have initial displacement θB0=0 rad, the initial velocity ωB0=16 rad/s and the angular acceleration is αB=-1.85 rad/s2-5.33t rad/s3+1.02t2 rad/s4. The first plot will be for the angular accelerations The above plot shows the angular acceleration of the two flywheels as a function of time. P is the point where the angular accelerations become equal. Now to find the angular displacement for flywheel A we have d2θdt2=1.90 Integrating once with respect to time we have ω=dθdt=1.90t+c1 Now the boundary condition gives ω0=-16 rad/s. Therefore c1=-16. Then dθdt=1.90t-16 Again integrating we get θ=1.902t2-16t+c2 Again the boundary condition we have θ0=0. This gives c2=0 Therefore for flywheel A we have θ=0.95t2-16t Now for the flywheel B we have d2θdt2=1.85-5.33t+1.02t2 Integrating once we get ω=dθdt=-1.85t-5.332t2+1.023t3+c1 Now we have the boundary condition ω0=16 rad/s which gives c1=16 Therefore dθdt=-1.85t-2.665t2+0.34t3+16 Again integrating we get θ=-1.852t2-2.6653t3+0.344t4+16t+c2 Again we have boundary conditions θ0=0. This gives c2=0. Thus θ=0.925t2-0.888t3+0.85t4+16t This graph shows the angular displacement vs time for the two flywheels.

Science

Advanced Physics

a) Sketch two graphs. The first graph should be angular acceleration versus time for the two wheel and clearly show the time at which the two angular accelerations are equal. The second graph should angular displacement as a function of time for the two wheels and should cover the first ten seconds of motion

a) Sketch two graphs. The first graph should be angular acceleration versus time for the two wheel and clearly show the time at which the two angular accelerations are equal. The second graph should angular displacement as a function of time for the two wheels and should cover the first ten seconds of motion