Question
Asked Dec 6, 2019
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A steel rod 1 meter long and having a mass of 2kg is pinned at one of its ends. A force is applied perpendicularly to the rod at a distance of 75 cm from the pinned end. What is the magnitude of the resulting angular acceleration of the rod? 
Irod=2kgm2

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Expert Answer

Step 1

Given values in the question:

Length of the steel rod, l = 1 m

Mass of the steel rod, m = 2 kg

Distance at which force is applied, r = 0.75 m

The expression for the moment of inertia I of the rod about one of the end is

 

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I =-ml 3 ( 2kg)(1m) = 0.67 kg m?

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Step 2

Consider the applied force as F.

The expression for the torque is,

 

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T = rF sin O

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Step 3

Angle between r and F is 90&d...

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T = 0.75F sin 90° = 0.75F

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