A light rod of length 1 m is pivoted at its center and two masses of 5

1.	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» 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 :-)*

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»

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 :-)

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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. ?

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