The above image shows a uniform slender bar hinged to move in the vertical plane

Question

The above image shows a uniform slender bar hinged to move in the vertical plane and pivoted about a fixed mechanism with a smooth pin A. The rod has a mass of 10 kg. A constant torque of T = 75 Nm is applied at the pin at time t = 0 s.

I. Calculate the resulting angular acceleration of the rod

II. Calculate the rotational speed (rpm) of the rod around pin A after t = 4s

III. Calculate the radial force acting on the pin A at t = 4s and the angle of the radial force with the horizontal axis. Illustrate it on a free-body-diagram.

IV. Calculate the kinetic energy of the rod at t = 4s and the work done by the torque T up to that time.

Please, keep the images at a reasonable resolution as Chegg downscales quite significantly. The working is needed as it's for finals revision!! Also, if all working could be shown it would be greatly appreciated too.

Explanation / Answer

moment of inertia of rod about hinge I=m*(1.5^3-0.5^3)/6=5.42 kgm^2 torque=moment of inertia*angular accelaration 75=5.42*alpha a).angular accelarion(alpha)=13.84 rad/s^2 b).assuming initial speed of rod to be zero(w1=0) w2=w1+alpha*t w2=13.84*4=55.38 rad/s c).total angle covered by rod in 4 sec=0.5*13.84*4*4=110.72radians=173.91 revolutions that means rod will rotate 327.6 degrees from its initial position total radial force acting on hinge is 15334.72N making angle of 32.4 with horizontal axis in downward direction d). total kinetic energy of rod=1/2 * I*w^2 8311.42 joules

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