1. In the figure below, two blocks, of mass m 1 = 300 g and m 2 = 710 g, are con

Question

1.In the figure below, two blocks, of mass m1 = 300 g and m2 = 710 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.

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In the figure below, two blocks, of mass m1 = 300 g and m2 = 710 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest. Find the magnitude of the acceleration of the blocks. Find the tension T1 in the cord at the left. Find the tension T2 in the cord at the right. The rigid body shown in the figure below consists of three particles connected by massless rods. It is to be rotated about an axis perpendicular to its plane through point P. If M = 0.37 kg,a = 25 cm, and b = 45 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s?

Explanation / Answer

Let the acceleration of m2 be a downwards

m2g - T2 = m2*a
0.71*9.8 - T2 = 0.71*a
so.. T2 = 6.958 - 0.71a

also ..
T1 - m1*g = m1*a
so.. T1 = 2.94 + 0.3a

for disk..

T2 - T1 = M*a/2 = 0.5 * a / 2
6.958 - 0.71a - 2.94 - 0.3a = 0.25*a

so. a = 3.18889 m/sec2


b) T1 = 2.94 + 0.3a = 3.89667 N

c) T2 = 6.958 - 0.71a = 4.693889 N




2)

inertia of the system about P = M*(b^2 - a^2) + 2*2M *a^2 = M*b^2 +3*M*a^2 = M*[ b^2 + 3*a^2 ]
= 0.37*[ 0.45^2 + 3* 0.25^2 ] = 0.1443 kg m^2

final angular speed = 5 rad/sec

so.. energy required = final kinetic energy = 0.5 * 0.1443 * 5^2 = 1.80375 j

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