The angular acceleration of the line L relative to the line

The angular acceleration of the line L relative to the line L_0 is given as a function of time by alpha = 2.5 - 1.2/rad/s. At t = 0. theta = 0 and the angular velocity of L relative to L_0 is omega = 5 rad/s. Determine theta and omega at t = 3 s. After solving the problem analytically, solve it again using MATLAB for numerical integration. Suggestion: Create a row vector of 301 times (t_n) from 0.00 s to 3.00 s. Create a row vector of angular accelerations a_n where each acceleration corresponds to the time according to alpha = 2.5 - 1.2t_n Using a loop on the column index, create a row vector of angular velocities omega_n. where the fust value is the initial angular velocity (5 rad/s), and at time t_n + 1 you compute omega_n + 1 = omega_n + alpha_n*(t_n+1 - t_n) Note that we are just using the definition of average acceleration: a_2= (omega_n + 1 - omega_n)/(t_n + 1 - t_n) Using a loop on the column index, create a row vector of angles theta_n, where the first value is the initial angle (0 rad), and at time t_n + 1 you compute theta_n + 1 = theta_n + omega_n*(t_n + 1 - t_n) How well do your numerically computed omega_301 and theta_301 match your analytical answers?

Solution

solution: 1) here we have to solve this problem by analytical as well by numerically computed by matlab code for numerical integration, so

2)for analytical approach

angular accelaration is rate of change of angular velocity,hence we can write

a=dw/dt=2.5-1.2*t

on integrating we get for t=0 to t=3

w=2.5*t-.6*t^2

for t=3 sec e get

w1=2.1 ard/s

but angular velocity is added to initial present hence total angular velocity is

w2=w1+w0

w2=2.1+5=7.1 rad/s

3) wher angular velocity is rate of change angle of position vector

w=d(angle)/dt

on integrating we get t=0 to t=3 sec

angle=1.25t^2-.2*t^3

for t=3 s

angle=5.85 rad

4) let solve it numerically in matlab code for trapezoidal rule

then we have

a=2.5-1.2t

here use three step n=3 and step size

h=tn-t0/n=3-0/3=1

hence t0=0

t1=1

t2=2

t3=3

and corresponding area

a0=2.5-1.2*0=2.5

a1=1.3

a2=.1

a4=-1.1

hence by trapezoidal rule area is

w1=A=h/2(a0+a3+2(a2+a3))=2.1

hence velocity is w1=2.1 rad/s

w2=w1+w0=5+2.1=7.1 rad/s

same process for angle subtended at center

w=2.5t-.6t^2

for same value step n=3 and step size h=1 sec

w0=2.5t-.6t^2=2.5*0-.6*0=0

w1=1.9

w2=2.6

w3=2.1

hence here by simpson 1/3 rule we have

angle=A=h/3(a0+a3+4(a1)+2(a2))=4.96 rad

6) here it is observe that angular velocity is same for both analytical and numerical approach but angle is different for both approach.

7)where row vector for accelaration ,velocity and angle is given as in term of time as

[a]=[-1.2][t]+[2.5]

[w]=[2.5][t]+[-.6][t^2]+[5]

[angle]=[1.25][t^2]+[-.2][t^3]