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3 years ago

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Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotationa

l inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

Explanation:

a )

Moment of inertial of four masses about axis that coincides with one side :

Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .

Moment of inertia of the two remaining masses

= m L² + m L²

= 2 mL²

b )

Axis that bisects two opposite sides

Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses

= 4 x m x ( L/2 )²

= 4 x mL² / 4

= m L² .

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Answer:

See Explanation

Explanation:

Given

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$s_0 \to L$ --- dimension

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Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

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$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

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Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

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Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

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Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

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$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

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$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

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$S_0 = \frac{24s}{t^4}$

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