Two boxes with masses m_1 and m_2 are attached via a pulley as shown, and the in

Question

Two boxes with masses m_1 and m_2 are attached via a pulley as shown, and the incline is frictionless and has angle Theta. The pulley is a uniform plate with moment of inertia I_p. For this problem, it may be useful to recall that y(t) = y_i + v_it + a_yt^2/2. Assume the system is initially at rest, compute the final velocity v_f of m_2 after it has fallen a distance of y. assume y, theta, I_p and the masses are all known. Please solve this part without finding the acceleration first. How much time t did it take the block to reach the v_f found in part (a)? Compute the frequency f of the allowed fundamental mode (n = 1) on the vertical segment of the string as a function of time t. Assume this segment initially has length L before the system was released, and string has linear mass density mu.

Explanation / Answer

(x2+9x) cosx dx

integration by parts :u =x2+9x =>du =2x+9 dx , dv =cosxdx =>v =sinx

udv =uv - vdu

(x2+9x) cosx dx =(x2+9x)sinx -sinx (2x+9) dx

(x2+9x) cosx dx =(x2+9x)sinx -9sinxdx -2xsinx dx

(x2+9x) cosx dx =(x2+9x)sinx +9cos -2xsinx dx

for integral part:

integration by parts : u =x , =>du =dx , dv =sinx dx , v =-cos x

udv =uv - vdu

(x2+9x) cosx dx =(x2+9x)sinx +9cosx -2[-xcosx --cosxdx]

(x2+9x) cosx dx =(x2+9x)sinx +9cosx -2[-xcosx +sinx] +c

(x2+9x) cosx dx =(x2+9x)sinx +9cosx +2xcosx -2sinx +c

(x2+9x) cosx dx =(x2+9x -2)sinx +(2x+9)cosx +c

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