Question

A thin rod pivoted at one end say 'O' and the rod is allowed to move in a horizontal circle, the length of rod is "L" which is of 2 m, a force of 4 N acts perpendicular to the length of the rod at free end produces torque of
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Answer 1

Answer:

Explanation:

At horizontal position the reaction 'N'.

Let the mass of rod is m.

The moment of inertia of rod about the axis passig through the end and perpendicular to its plane is

I=  

3

1

​  

ml  

2

 

For rotation of rod,

Moment of force = Applied torque

⟹x×F=Iα

⟹α=  

3

1

​  

ml  

2

 

xF

​  

=  

ml  

2

 

3xF

​  

 

Here α is angular acceleration about hinge .

The position of centre of mass  is aat the middle.

So the linear acceleration of centre of mass,a=α  

2

l

​  

 

a=  

2ml

3xF

​  

 

So, the net force in vertical on centre of mass is  

=ma

=  

2l

3xF

​  

 

This should be equal to N.

So, N=  

2l

3xF

​