Question

Four particles, each of mass 0.16 kg are placed at the vertices of a square with sides of length 0.42 m The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal, as shown in the figure.(A) What is the rotational inertia of the body about axis A?(B) What is the angular speed of the body about axis A at the instant rod AB swings through the vertical position?

Answer 1

Answer:

0.112896 kgm²

6.83478497937 rad/s

Explanation:

m = Mass of each particle = 0.16 kg

l = Length of side = 0.42 m

$\omega$ = Angular speed

g = Acceleration due to gravity = 9.81 m/s²

Moment of inertia would be given by

$I=0+ml^2+m(\sqrt2 l)^2+ml^2\\\Rightarrow I=ml^2+m2l^2+ml^2\\\Rightarrow I=4ml^2\\\Rightarrow I=4\times 0.16\times 0.42^2\\\Rightarrow I=0.112896\ kgm^2$

Rotational inertia of the body about axis A is 0.112896 kgm²

Applying conservation of energy

$4mgl=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{4mgl\times 2}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{4\times 0.16\times 9.81\times 0.42\times 2}{0.112896}}\\\Rightarrow \omega=6.83478497937\ rad/s$

The angular speed is 6.83478497937 rad/s

Answer 2

A) The rotational inertia of the body about axis A = 0.113 kgm²

B) The angular speed of the body about axis A = 6.83 rad/s

Given data :

number of particle = 4

mass of each particle ( m ) = 0.16 kg

length of side ( L )  = 0.42 m

g ( acceleration due to gravity ) = 9.81 m/s²

A) calculate the rotational inertia of the body about axis A

moment of inertia ( I ) = 0 + mL² + m (√2l )² + mL²

∴ Rotational moment of inertia ( I ) = 4mL²

                                        = 4 * 0.16 * 0.42² ≈ 0.113 kgm²

B) Determine the angular speed of the body

4mgI = $\frac{1}{2}Iw^{2}$  ( conservation of energy )  ------ ( 1 )

making  w  the subject of the equation to determine the angular speed

w = $\sqrt{\frac{4mgl*2}{I} }$  ------- ( 2 )

where ; m = 0.16kg,  g = 9.81 m/s²,   L = 0.42 m,  I = 0.113 kgm²

Insert values into equation 2 above

w ≈ 6.83 rad/s

Hence we can conclude that the  rotational inertia of the body about axis A = 0.113 kgm² and The angular speed of the body about axis A = 6.83 rad/s.

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