Question

Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. The rod is initially held in horizontal position but eventually allowed to swing down. 1.2 m 3.2 kg O 38◦ What is its angular acceleration at the instant it makes an angle 38 ◦ with the horizontal?

Answer 1

Answer:

the angular acceleration is 9.7 rad/$s^{2}$

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

τ = torque

I = inertia

σ = acceleration

moment inertia of this rod, I

I = $\frac{1}{3} mL^{2}$

τ = F d, d = $\frac{L}{2}$cosθ

τ = m g $\frac{L}{2}$cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

  = (m g $\frac{L}{2}$cosθ)/($\frac{1}{3} mL^{2}$)

  = 3gcosθ/2L

  = 3 (9.8)cos 38°/(2 x 1.2)

  = 9.7 rad/$s^{2}$