# Angular Acceleration of a Fixed Rod

**1. The problem statement, all variables and given/known data**

Consider a thin 14 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 3.2 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is Irod =1/3mL^2 and the moment of inertia of the sphere about its center of mass is Isphere =2/5mr^2 . What is the angular acceleration of the rod immediately after it is released from its initial position of 35◦ from the vertical? The acceleration of gravity g = 9.8 m/s^2

If needed a graphical representation of the question can be found at the following link:

http://ift.tt/OpPkbU

**2. Relevant equations**

τ = Iα

I = (1/3)MR^{2}

τ = FRsin(Θ)

**3. The attempt at a solution**

Finding the position of the center of mass (x meters from pivot):

(x-7) = (14+3.2-x)

x=12.1 meters

Using x_{cm} = 12.1 m, we can solve for α:

I = (1/3)(18 kg)(12.1 m)^{2} ≈ 878.46 kg·m^{2}

τ = (18 kg)(9.8 m/s^{2})(12.1 m)sin(35°) ≈ 1224.264489 N·m

α = τ/I ≈ 1.39361 rad/s^{2}

This answer is taken as incorrect by my online program. Any suggestions or assumptions that I have made in my calculations would be helpful. Thanks. One thing that I was worried about was that this method didn’t use the moments of inertia that I was given.

http://ift.tt/1hka7dP

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