A spool of mass ms - 150 kg and inner and outer radii p - 0.8m and R - 1.2 m, re

Question

A spool of mass ms - 150 kg and inner and outer radii p - 0.8m and R - 1.2 m, respectively, is connected to a counterweight A of mass mA - 50 kg by a pulley system whose cord, at one end, is wound around the inner hub of the spool. The center G of the spool is also the center of mass of the spool, and the radius of gyration of the spool is kG = 1 m. The system is at rest when the counterweight is released, causing the spool to move to the right. Assume that the spool rolls without slip. Neglecting the inertia of the pulley system, determine the angular speed of the spool after the counterweight has dropped 0.5 m. Assume that the inertia of the cord and of pulleys B and D can be neglected, but model pulley C as a uniform disk of mass mC - 15 kg and radius rC - 0.3 m. If the cord does not slip relative to pulley C, determine the angular speed of the spool after A drops 0.5 m.

Explanation / Answer

Let the angular acceleration of the spool be aplha
linear acceleraion of spool = as = alpha * R = 1.2*alpha
acceleration of rope w.r.t top point of cylinder = rho*alpha = 0.8*alpha

so.. total acceleration of the rope (a)= 1.2*alpha + 0.8 alpha = 2 * alpha

....Linear acceleriaon of spool = 1.2*alpha

now.. acceleration of counterweight = (1/3)*acceleraion of rope from spool
so.. a_A = 2*alpha / 3

Let the tension in the rope be T and the friction between spool & ground be f

for spool ...
T - f = M_s*a_s
so.. T - f = 150*1.2*alpha
so.. T - f = 180 * alpha
so.. f = T - 180 alpha

also... for intertia
T*rho + f*R = M_s * k^2 * alpha
0.8T + 1.2f = 150*1^2 * alpha
so.. 0.8T + 1.2*(T - 180 alpha) = 150 alpha
so.. 2T - 216 alpha = 150 alpha
so... T = 183 alpha

for counterblock ..

mA*g - 3*T = mA*a_A
50*9.81 - 3*183*alpha = 50*2*alpha / 3
so.. alpha = 50*9.81 / ( 582.3333) = 0.842301 rad/sec2

so.. acceleraion of counter weight = a_A = 2*alpha / 3 = 2 * 0.842301 / 3
so.. a_A = 0.561534 m/sec2

now... for counterweight to fall 0.5 m ..
distance s = 0.5 m
acceleraion =a = 0.561534 m/sec2
inital velocity u = 0
let time = t

s = ut + 0.5 * a*t^2
0.5 = 0 + 0.5*0.561534*t^2

so.. time taken = t = 1.33448 sec

so... angular speed of spool after 1.33448 sec
= time * alpha = 1.33448 * 0.842301 = 1.124034 rad/sec

Problem . 8.47

for spool ...
T - f = M_s*a_s
so.. T - f = 150*1.2*alpha
so.. T - f = 180 * alpha
so.. f = T - 180 alpha

also... for intertia
T*rho + f*R = M_s * k^2 * alpha
0.8T + 1.2f = 150*1^2 * alpha
so.. 0.8T + 1.2*(T - 180 alpha) = 150 alpha
so.. 2T - 216 alpha = 150 alpha
so... T = 183 alpha

let the tension in rope from C to A be T2

for disk C
inertia = m_C*r_C^2 / 2 = 15*0.3^2/2 = 0.675 kg m^2
angular acceleraion of pulley = alpha_C = acceleration of rope / R_C = 2*alpha / 0.3 = 6.666667 * alpha

so.. for pulley.
(T2 - T ) * R_C = inertia * alpha_C
(T2 - 183 alpha ) * 0.3 = 0.675 * 6.666667 * alpha
so.. T2 = 198*alpha

for counterblock ..

mA*g - 3*T2 = mA*a_A
50*9.81 - 3*(198*alpha) = 50*2*alpha / 3
so.. alpha = 50*9.81 / ( 627.3333) = 0.78188 rad/sec2

so.. acceleraion of counter weight = a_A = 2*alpha / 3 = 2 * 0.78188 / 3
so.. a_A = 0.521253 m/sec2

now... for counterweight to fall 0.5 m ..
distance s = 0.5 m
acceleraion =a = 0.521253 m/sec2
inital velocity u = 0
let time = t

s = ut + 0.5 * a*t^2
0.5 = 0 + 0.5*0.521253*t^2

so.. time taken = t = 1.385083 sec

so... angular speed of spool after 1.33448 sec
= time * alpha = 1.385083 * 0.78188 = 1.082969 rad/sec

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