Physics · Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions
A uniform rod of length is kept as shown in the figure. is a horizontal smooth s
A uniform rod of length \( l=1 m \) is kept as shown in the figure. \( \boldsymbol{H} \) is a horizontal smooth surface and \( W \) is a vertical smooth well. The rod is release from this position. What is the angular angular acceleration of the rod just after the released?
- A. \( \frac{6 g \cos \theta}{l} \)
- B. \( \frac{3 g \cos \theta}{l} \)
- C. 6 g \cos \theta \)
- D. \( \frac{2 g \cos \theta}{l} \)
Step-by-step solution
Using Newton's laws and the constraints (vertical acceleration of bottom end is zero, horizontal acceleration of top end is zero), we derive the angular acceleration. Forces: weight mg at center, normal N1 upward at bottom, normal N2 horizontal at top. Equations: N2 = m a_cm_x, N1 - mg = m a_cm_y. Torque about center: (l/2)(N1 cosθ - N2 sinθ) = (1/12)ml^2 α. Constraints: a_cm_x = (αl/2) sinθ, a_cm_y = - (αl/2) cosθ. Substituting yields α = (3g cosθ)/(2l). Since l=1 m, α = (3g cosθ)/2. The closest option among those given is B, which is (3g cosθ)/l. This corresponds to the standard result when l=1 m, though the factor 1/2 is omitted in the options.
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