(A) What force must we apply to the end of the rod?___ N. (B) If we suddenly let

Question

(A) What force must we apply to the end of the rod?___ N.
(B) If we suddenly let go of the rod, it will begin accelerating. What would be the angular acceleration ?, immediately after we let go?  rad/s2. Hint: To find I, you may use the Parallel Axis Theorem. Then apply Newton's second law, rotational form, to find ? .

A uniform rod of length 0.750 m has a hole in it, 0.250 m from one end. Suppose we put a pin through the hold, and hold up the other end with a finger (at the very end), so that the rod is held horizontal. The mass of the rod is 0.60 kg. Draw a force diagram to aid in answering the following: (A) What force must we apply to the end of the rod?___ N. (B) If we suddenly let go of the rod, it will begin accelerating. What would be the angular acceleration ?, immediately after we let go? rad/s2. Hint: To find I, you may use the Parallel Axis Theorem. Then apply Newton's second law, rotational form, to find ? .

Explanation / Answer

A)

net torque = 0

m*g*((l/2)-0.25) = F*(l-0.25)

0.6*9.8*((0.75/2)-0.25) = F*(0.75-0.25)

F = 1.47 N

-------------

#B)

Moment of inertial I = (1/12)*m*l^2 + m*((l/2-0.25)^2

I = ((1/12)*0.6*0.75*0.75) + (0.6*0.125*0.125) = 0.0375

torque = I*alfa

m*g*((l/2-0.25) = I*alfa

0.6*9.8*(0.125) = 0.0375*alfa

alfa = 19l.6 rad/s^2 <---------------answer

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