Sir
Pl solve Ques no -11
Asked by shobhit
23rd June 2018,
12:00 AM
Answered by Expert
Answer:
Once the plate is released from the right edge (point B), it rotates around the point A.
Torque due to the force F = mg is given by
...................(1)where r is the vector and its magnitude is the distance between A and centre of mass. I is moment of Inertia.
α is angular acceleration
distance r can be calculated from given dimensions.
r = (√5/4)c
r × mg = r×mg×sinθ = r×mg×0.447 ...............(2)
in eqn.(2), θ is calculated as 26.57º from the given dimension(refer figure)
moment of inertia I 
.................(3)
.................(3)from eqns.(1), (2) and (3), we get α = 5.877/C rad/s
acceleration of CM = r×α = (√5/4)c × ( 5.877 / c) = 3.285 m/s
Answered by Expert
29th June 2018,
6:04 PM
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