A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 2.65 m and mass 6.84 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.301 rad/s2
We tried T = mr^2(alpha), but that was wrong. We also tried using the length of the whole ladder instead of the radius, and again, incorrect
Well, you missed the moment of inertia for a thin rod be a factor of 12.
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/RotationalKinematics/MomentInertia.html
Thanks man!
You're welcome! I'm here to help.
To find the torque the person must exert on the ladder, we need to use the correct formula for moment of inertia. The moment of inertia for a thin rod rotating about one end is given by the equation I = (1/3) * m * L^2, where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.
In this case, the length of the ladder is given as 2.65 m, and the mass is given as 6.84 kg. So, we can substitute these values into the equation to find the moment of inertia:
I = (1/3) * 6.84 kg * (2.65 m)^2 = 6.91 kg⋅m^2
Now that we have the moment of inertia, we can use the equation for torque, τ = I * α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Given that α is 0.301 rad/s^2 from the problem, we can substitute the values into the equation to find the torque:
τ = 6.91 kg⋅m^2 * 0.301 rad/s^2 = 2.082 N⋅m
Therefore, the person must exert a torque of 2.082 N⋅m on the ladder to give it an angular acceleration of 0.301 rad/s^2.
I hope this explanation helps! Let me know if you have any further questions.