Answer:
4.535 N.m
Explanation:
To solve this question, we're going to use the formula for moment of inertia
I = mL²/12
Where
I = moment of inertia
m = mass of the ladder, 7.98 kg
L = length of the ladder, 4.15 m
On solving we have
I = 7.98 * (4.15)² / 12
I = (7.98 * 17.2225) / 12
I = 137.44 / 12
I = 11.45 kg·m²
That is the moment of inertia about the center.
Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that
τ = 11.453 kg·m² * 0.395 rad/s²
τ = 4.535 N·m
The Torque will be "4.535 N.m".
Given:
Mass of ladder,
Length of ladder,
The moment of inertia will be:
→ [tex]I = \frac{mL^2}{12}[/tex]
[tex]= \frac{7.98\times (4.15)^2}{12}[/tex]
[tex]= \frac{7.98\times 17.2225}{12}[/tex]
[tex]= 11.45 \ kg.m^2[/tex]
hence,
The torque will be:
→ [tex]\tau = 11.453\times 0.395[/tex]
[tex]= 4.535 \ N/m[/tex]
Thus the above approach is correct.
Learn more about torque here:
https://brainly.com/question/19247046
