Question

"The slender rod AB shown has a mass of m=63.0 kg and is being supported by a rope and pulley system stationed at C. Starting from rest (in the position shown), the rope and pulley system tug on the rod causing it to rotate about A. The torque applied to the pulley is T=2.85 kN⋅m and has an effective moment arm of r=0.160 m. The dimensions shown in the figure are l=2.90 m and h=1.10 m. Assume the pulley is frictionless and massless. Determine the angular acceleration of the rod the instant the rope and pulley system have pulled the rod through an angle of θ=2.50∘."

Answer 1

I've found Beta = 21.1 deg. and F = 17.8 kN, because the hint system instructed me to. All I need to know is what equation of motion I need to find alpha, and then I'll be able o solve the rest of the problem myself.

Answer 2

i'm not so sure it's that simple.

check this out and see what you think...

the aim being to use \(\tau = I \alpha \), firstly the torque from the rod itself about A is just \(\tau_{rod} = \dfrac{1}{2}mgl \cos \theta \) where \(\theta\) is the angle the rod is above the horizontal

the pulley system is a bit of a nuisance and can be replace by a tension in the string \(T_c\) pulling toward C, where \(T_c = 2.85 kNm \times 0.16m\)

the torque about A of that tension is \(\tau_A = \vec r \times \vec T_c\), where:

\(\vec r = \vec{AB} = < l \cos \theta, \; l \sin \theta>\)

direction of \(\vec T_c\) is \(\vec {BC} = \vec {AC} -\vec {AB} \)

or \(<0,h>- < l \cos \theta, \; l \sin \theta> = <- l \cos \theta, \;h- l \sin \theta>\)

so \(\hat{T_c} = \dfrac{<- l \cos \theta, \;h- l \sin \theta>}{\sqrt{l^2 \cos^2 \theta + h^2 - 2hl \sin \theta + l^2 \sin^2 \theta }} \)


\(=\dfrac{<- l \cos \theta, \;h- l \sin \theta>}{\sqrt{l^2 + h^2 - 2hl \sin \theta }} \)

so \(\tau_A = \vec r \times \vec T_c\)

\(= < l \cos \theta, \; l \sin \theta> \times \dfrac{<- l \cos \theta, \;h- l \sin \theta>}{\sqrt{l^2 + h^2 - 2hl \sin \theta }} \color{red}{\times [ 2.85 kNm \times 0.16m]} \)

\(=\dfrac{ 2.85 kNm \times 0.16m }{\sqrt{l^2 + h^2 - 2hl \sin \theta }} \; \left|\begin{matrix} l \cos \theta & l \sin \theta \\- l \cos \theta & h-l \sin \theta\end{matrix}\right|\)

\(=\dfrac{ (2.85 kNm \times 0.16m) \; lh \cos \theta }{\sqrt{l^2 + h^2 - 2hl \sin \theta }} \)

and from \(I \alpha = \tau_A - \tau_{rod} \)

\( \dfrac{1}{3}ml^2 \ddot \theta = \dfrac{ (2.85 kNm \times 0.16m) \; lh \cos \theta }{\sqrt{l^2 + h^2 - 2hl \sin \theta }} - \dfrac{1}{2}mgl \cos \theta \)

note you can actually integrate that quite easily to \( \dfrac{1}{6}ml^2 \dot \theta^2 = - (2.85 kNm \times 0.16m)\sqrt{l^2 + h^2 - 2hl \sin \theta } - \dfrac{1}{2}mgl \sin \theta + C \)
which suggests it is along the right lines

well, that's in outline how i would try and do it, FWIW

Answer 3

incidentally that minus sign in the energy equation worries me slightly, there might be a typo in the algebra.....