In the figure shown, the rod OA of length 1 m is rotating about the fixed point O and the rod AB of length 1 m is rotating about A, in a vertical plane. At a certain moment, when OA is making an angle 60° with the horizontal, the rod AB is horizontal. At the same moment angular speed and angular acceleration of OA and AB are ωₒₐ = 2 rad/s, αₒₐ = 2√3 rad/s², ωₐᵦ = 1 rad/s, αₐᵦ = 3√3 rad/s² respectively. Calculate the acceleration of point B (in m/s²) at that moment.
Physics · Mixed Chapter
JEE Mains

Question

In the figure shown, the rod OA of length 1 m is rotating about the fixed point O and the rod AB of length 1 m is rotating about A, in a vertical plane. At a certain moment, when OA is making an angle 60° with the horizontal, the rod AB is horizontal. At the same moment angular speed and angular acceleration of OA and AB are ωₒₐ = 2 rad/s, αₒₐ = 2√3 rad/s², ωₐᵦ = 1 rad/s, αₐᵦ = 3√3 rad/s² respectively. Calculate the acceleration of point B (in m/s²) at that moment.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The acceleration of B is the vector sum a_B=a_A + α_AB×r_BA − ω_AB²r_BA. Step 2: Similarly a_A=α_OA×r_AO−ω_OA²r_AO. Resolve these expressions into x and y components using r_AO=(cos60°, sin60°) m and r_BA=(1,0) m. Step 3: The signs of both angular accelerations depend on the curved arrows in the original figure. For the signs indicated by the drawing, check whether the requested value is the x-component or the magnitude: the printed answer is 0 but the net vector can have a non-zero y-c…