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n200080 [17]
3 years ago
13

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.
Physics
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:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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