The angular joint velocity, V (deg/s), of a 2-link robot arm is given by the fol

Question

The angular joint velocity, V (deg/s), of a 2-link robot arm is given by the following equation where t is time in seconds and Tf (seconds) is the total time required to move from the starting position to the final position. V = 30/t^5_f t^4 - 60/T^4_f t^3 + 30/T^3_f t^2 Find all of the times, t, at which the joint velocity is either maximized or minimized. Note: some of the critical points will be a function of the unknown constant T_f. Critical Points (t-values):____________ MATLAB Commands and/or Work: Apply the 2^nd derivaties test to determine the time(s) at which velocity is maximized and the time(s) at which velocity is minimized. Time(s) at which velocity is maximized: Time(s) at which velocity is minimized: MATLAB Commands and/or Work: Suppose Tf = 6 seconds. Find the maximum and minimum joint velocity. Maximum Velocity:___________ Minimum Velocity:__________ MATLAB Commands and/or Work:

Explanation / Answer

1)answer

syms t

syms x

v=((30/x^5)*t^4)-((30/x^4)*t^3)+((30/x^3)*t^2);

diff(v);

crit_pts = solve(k);

fprintf('the critical points are');

disp(crit_pts);

Output::

the critical points are

t/2 + (3^(1/2)*t*i)/2

t/2 - (3^(1/2)*t*i)/2

2)answer

syms t

syms x

v=((30/x^5)*t^4)-((30/x^4)*t^3)+((30/x^3)*t^2);

% first derivative

g=diff(v);

    disp(g);

    firstderivate = simplify(g)

    % second derivative

secondderivative = diff(firstderivate)

%   To find the inflection point of f, set the second derivative equal to 0 and solve.

inflec_pt = solve(f2,'MaxDegree',2)

disp(inflec_pt)

% after calculating the inflection point the equation are obtained

output:::

imnmmm

(120*t^3)/x^5 - (90*t^2)/x^4 - (150*t^4)/x^6

firstderivate =

-(30*t^2*(5*t^2 - 4*t*x + 3*x^2))/x^6

secondderivative =

(30*t^2*(4*t - 6*x))/x^6 + (180*t^2*(5*t^2 - 4*t*x + 3*x^2))/x^7

inflec_pt =

(5*t)/6 + (65^(1/2)*t*i)/6

(5*t)/6 - (65^(1/2)*t*i)/6

(5*t)/6 + (65^(1/2)*t*i)/6

(5*t)/6 - (65^(1/2)*t*i)/6

The maximum and minimum values are 0

anotheraway

syms t

syms x

v=((30/x^5)*t^4)-((30/x^4)*t^3)+((30/x^3)*t^2);

d=diff(v);

g=diff(d);

maxandminpoints = solve(g);

fprintf(maxandminpoints);

disp(maxandminpoints);

output::

(5*t)/6 + (65^(1/2)*t*i)/6

(5*t)/6 - (65^(1/2)*t*i)/6

If sove this equations than answer is 0 obtained

3)answer

Here x means Tf for simplicatins I put x= Tf

x=6;

k=((30/x^5)*t^4)-((30/x^4)*t^3)+((30/x^3)*t^2);

n=solve(k);

disp(n)

maxpoint=max(n);

fprintf('the max points are');

dsp(maxpoint);

minpoint=min(n);

fprintf('the min points are');

disp(minpoint);

output::

the max points are 3.0000 + 5.1962i

the min points 0

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