A 8.78 -m ladder with a mass of 23.8 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 236 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.90 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?
About what rotation axis do you want the moment of inertia? You need to specify it. Is it at one of the ends? The center?
With the ladder laying flat on the ground, the axis is on the left and it is being pulled upward on the right. But the center of gravity confuses me. I don't know how that comes into play. These topics are still new to me.
See my answer to a similar question below under related questions.
To find the net torque acting on the ladder, we need to use the formula:
Net Torque = Moment of Inertia * Angular Acceleration
To find the moment of inertia of the ladder, we can use the formula:
Moment of Inertia = Mass * Length^2 / 3
Let's calculate each part:
(a) Net Torque:
Given: Angular Acceleration (α) = 1.90 rad/s^2
Using the formula Net Torque = Moment of Inertia * Angular Acceleration, we need to calculate the moment of inertia.
(b) Moment of Inertia:
Given: Mass (m) = 23.8 kg
Length (L) = 8.78 m
Using the formula Moment of Inertia = Mass * Length^2 / 3, we can calculate the moment of inertia.
Now let's calculate:
(a) Net Torque:
Net Torque = Moment of Inertia * Angular Acceleration
(b) Moment of Inertia:
Moment of Inertia = Mass * Length^2 / 3
Let's plug in the given values and calculate the answers.