A block of mass \"6m\" is tied to another block of mass \"m\" after passing over

Question

A block of mass "6m" is tied to another block of mass "m" after passing over a frictionless pulley (Disk I = 1/2 mR^2)of mass "m" and radius "R".The system starts from rest. The final position is after the "6m" block drops a distance h. a.Write out the energy equation for the system between the initial and final positions. Given [m, R, h] Determine b.The speed of the blocks at the final position. b.The angular speed of the disk at the final position. c.The acceleration of the blocks. e.The tension in the string above the 6m block. f.The tension in the string above the m block. g.The time for the "6m" block to fall the distance h.

Explanation / Answer

a) P. Energy lost by 6m mass = 6mgh
PE gained by m mass = mgh
KE gained by 6m mass = 0.5*6m*v^2
KE gained by m mass = 0.5mv^2
Rotational KE gained by pulley = Iw^2 = 0.5*mR^2*v^2/R^2 = 0.5mv^2
SO, by energy balance
6mgh = mgh + 3mv^2 + 0.5mv^2 + 0.5mv^2 = mgh + 4mv^2
b) 5mgh = 4mv^2
5gh = 4v^2
v = sqroot(5gh/4)
c) w = v/R = sqroot(5gh/4)/R
d) acceleration of blocks = a
T1 - mg = ma
6mg - T2 = 6m*a
[T1 - T2] + 5mg = 7*ma

[T2 - T1]R = 0.5mR^2*a/R
[T2 - T1] = 0.5m*a
a = 2[T2 - T1]/m = 2[7ma - 5mg]/m = 2[7a - 5g]
a = 14a - 10g
10g = 13a
a = 10g/13

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