Question

A light rod of length 1 m is pivoted at its center and two masses of 5 kg and 2 kg are hung from the ends as shown in the figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning. ?

Answer 1

Given in the question :-

Length l = 1 m

Mass m₁ = 2 kg

m₂ = 5 kg.

Now we know the formula

F₁ r₁ - F₂r₂ = m₁r₁² + m₂r₂²

Hence,

$I = 2 * 0.5^2+ 5 * 0.5^2$

I = 7 × 0.25

I =1.75 kg.meter² .

Torque about the pivot, i.e T = m₁r₁ + m₂r₂

T = 5 × 0.5 - 2 × 0.5

T = (1.5 × g) Nm

Therefore the acceleration

$\alpha = \frac{T}{I}$

α = (1.5 × g) / 1.75

α = (1.5 × 9.8)/ 1.75

α = 14.7 / 1.75

α = 8.4 rad/second²

Hope it Helps :-)

Answer 2

●Total moment of Inertia of the system about the axis rotation.

$= > I _{net} = (m _{1} {r _{1} }^{2} + m _{2} {r _{2} }^{2} ) \\ \\ = > TORQU = T _{net} = F _{1}r _{1} - F _{2}r _{2} \\ \\ also \\ \\ = > T _{net} = I _{net} \times a$
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●On equating the value of Tnet and putting the value of Inet, we get :-

$= > F _{1}r _{1} - F _{2} r _{2} = (m _{1} {r _{1} }^{2} + m _{2} {r _{2} }^{2}) \times a \\ \\ = > ( - 2 \times 10 \times 0.5) + (5 \times 10 \times 0.5) \\ = (5 \times { \frac{1}{2} }^{2} + 2 \times { \frac{1}{2} }^{2} )a \\ \\ = > 15 = \frac{7}{4} a \\ \\ = > a = \frac{60}{7} \\ \\ = > a = 8.57 \: \: \frac{rad}{ {s}^{2} }$
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