A 2.50 Kg square metal plate 0.180m on each side is pivoted about an axis throug

Question

A 2.50 Kg square metal plate 0.180m on each side is pivoted about an axis through point at its center perpendicular to the plate as in figure below. The magnitude of the forces are F_1 = 18.0 N, F_2 = 26.0 N, and F_3 = 14.0 N and the plate and all forces are in the plane of the page. the plate is initially at rest. a. Calculate the net torque about this axis due to the three forces shown in the figure. b. Find the angular acceleration of the plate. (I_plate = mL^2/6) c. If the plate keeps rotation with the angular acceleration you found in part (b), find angular speed and the total angle the plate rotates after Delta t = 3.00 s.

Explanation / Answer

T1 = -18*0.09 = -1.62 Nm

T2 = 26*0.09 = 2.34 Nm

T3 = 14*0.09/1.414 = 0.89 Nm

T = 3.23-1.62 = 1.61 Nm

T = I*alpha

alpha = T/I

I = 2.5*0.18^2/6 = 0.0135 kg m^2

alpha = 1.61/0.0135 = 119.26 rad/s^2

c) angular speed = alpha/t

w = 119.26/3 = 39.75 rad/sec

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