Question

The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismoved slightly to the right and then released, determine the veloc-360 mimity of pin B after rod AB has rotated through 90*.

Answer 1

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

What is Potential and Kinetic energy?

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

$T_{1} + V_{1} = T_{2} + V_{2}$

$h_{ab} = h_{bc} = 0.18m$

potential energy at position 1,

$V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}$

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

$T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}$

$l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}$

= 1/3 *4 * (0.36)²

=0.10368kg m²

$l =\frac{m_{bc} l^{2}_{bc} }{12}$

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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