A9.76-m ladder with a mass of 22.6 kg lies flat on the ground. a painter grabs the top end of the ladder and pulls straight upward with a force of 254 n. at the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.99 rad/s2 about an axis passing through the bottom end of the ladder. the ladder's center of gravity lies halfway between the top and bottom ends.
a) what is the net torque acting on the ladder?
b) what is the ladder's moment of inertia?

The definition of torque and Newton's second law for rotation allows to find the results for the questions about the rotational motion of the ladder are:

   A) The net torque is:  τ = 1397.1 N m

   B) the moment of inertia of the ladder with respect to the pivot point is:

        I = 702.1 Kg m²

Torque is the product of force for the distance perpendicular to the point of application and is the effect of force on angular motion.

        τ = F x r

Where bold indicates vectors, τ is torque, F is force, and r is distance to the pivot point.

The module of this expression can be written explicitly

           τ = F r sin θ

where the quantity (r sin θ) is called the perpendicular distance or arm.

Part A)

In this case they indicate that the applied force is F = 254 N at the ends of the ladder at L = 9.76 m from the turning point and the center of mass is at the middle point of the ladder.

In the attached we can see a diagram of the forces, where the reference system at the turning point and the counterclockwise as positive.

        ∑τ = F L - W L / 2

Weight is:   W = mg

        Σ τ = F L - mg L / 2

        Σ τ = 254 9.76 - 22.6 9.8 9.76 / 2

        Σ τ = 1397.1 Nm

Part B)

Newton's second law for rotational motion establishes the relationship between the net torque and the product of the moment of inertia and the angular acceleration.

       Σ τ = I α

They indicate the value of the angular acceleration α= 1.99 rad/a²

       I =  

Let's calculate

      I =  

      I = 702.1 kg m²

In conclusion using the definition of torque and Newton's second law for rotation we can find the results for the questions about the rotational motion of the ladder are:  

   A) The net torque is τ = 1397.1 Nm

   B) The moment of inertia of the ladder with respect to the pivot point is:

         I = 702.1 Kg m²

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Part a)

Part b)

Explanation:

Part a)

Torque on the rod is due to two forces

(i) Due to weight of the rod at mid point

(ii) Due to external force applied at the end

Now we have

Part b)

As per Newton's law we know that

now we have

wavelength variability is the right one

v=ir = 3x15 =45 v