A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating

Question

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4 rpm. What is its kinetic energy As shown in the figure, two blocks are connected by a light string that passes over a frictionless pulley having a moment of inertia of 0.0040 kg. m^2 and diameter 10 cm. The coefficient of kinetic friction between the table top and the upper block is 0.30. The blocks are released from rest, and the string does not slip on the pulley. How fast is the upper block moving when the lower one has fallen 0.60 m The figure shows two blocks connected by a light cord over a pulley. This apparatus is known as an Atwood's machine. There is no slipping between the cord and the surface of the pulley. The pulley itself has negligible friction and it has a radius of 0.12 m and a mass of 10.3 kg. We can model this pulley as a solid uniform disk. At the instant that the heavier block has descended 15 m starting from rest, what is the speed of the lighter block

Explanation / Answer

17)
r = 10 cm
m = 3.0 Kg
w = 33.4 rpm = 33.4 * (2*)/60
w = 3.50 rad/s

K.E = 1/2 * I * w^2
K.E = 1/2 * 1/2 * m*r^2 * w^2
K.E = 1/4 * 3.0 * 0.1^2 * 3.5^2
K.E = 0.0918 J


2.0 *g - T1 = 2.0*a
T1 = 2.0*g - 2.0 * a

T2 - 0.3*3.0*g = 3.0 * a
T2 = 3.0*a + 0.9*g

(T1 - T2)*r = I *a/r
(2.0*g - 2.0 * a - 3.0*a - 0.9*g ) = I*a/r^2
(1.1 * 9.8 - 5.0*a) = 0.0040/(0.05)^2 * a
a = 1.63 m/s^2

Now,
v^2 = u^2 + 2*a*s
v^2 = 0 + 2*1.63*0.6
v = 1.39 m/s
Speed of upper block, v = 1.39 m/s

5.7*g - T1 = 5.7 * a
T2 - 3.0*g = 3.0*a

(T1 - T2)*r = I *a/r
(5.7*g - 5.7*a - 3.0*a - 3.0*g) = I*a/r^2
(2.7 * 9.8 - 8.7 * a) = 1/2*10.3 * a
a = 1.91 m/s^2

v^2 = u^2 + 2*a*s
v^2 = 0 + 2*1.91*1.5
v = 2.39 m/s

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