A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot in the vertical plane. The rod is held in the horizontal position and released from rest. What is the initial angular acceleration of the rod and the initial linear acceleration of the right end of the rod?

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-01-05T10:38:01+01:00

The angular acceleration is 32.67 rad per sec squared and the initial linear acceleration of the right end of the rod is 29.4 meters per second squared

Torque- A twisting force cause of rotation is known as torque. It is the product of the force and perpendicular distance of the line of action of a force from the axis of rotation. Mathamatically,

[tex]\tau =mg \times sin\theta\times l[/tex]

As tod is held in the horizontal position,

[tex]\tau =mgl[/tex]

Torque is the product of angular acceleration and moment of inertia. Mathamatically,

[tex]\tau =\alpha I[/tex]

To evaluate the value of angular acceleration equate both the values of the torque.

[tex]\alpha I=mgl[/tex]

Moment of inertia can be written as,

[tex]I=\dfrac{ml^2}{3}[/tex]

Putting this value in the above equation,

[tex]\alpha \dfrac{ml^2}{3} =mgl[/tex]

[tex]\alpha =\dfrac{3mgl}{ml^2}[/tex]

[tex]\alpha =\dfrac{3g}{l}[/tex]

[tex]\alpha =\dfrac{3\times 9.8}{0.9}[/tex]

[tex]\alpha =32.67[/tex]

Hence, the angular acceleration is 32.67 rad per sec squared

At the initial point, the linear acceleration is tangential, therefore,

[tex]\alpha _{l} =\alpha l[/tex]

[tex]\alpha _{l} =32.67\times0.9[/tex]

[tex]\alpha _{l} =29.4[/tex]

Hence, the initial linear acceleration of the right end of the rod is 29.4 meters per second squared

For more about the torque, follow the link below-

https://brainly.com/question/6855614