Question

17.
A thin wire of length_l and mass m is bent in
the form of a semicircle as shown. Its moment of
inertia about an axis joining its free ends will
be :
​

Answer 1

Answer:

Explanation:

Length = Circumference of semicircle

l=nA

r=  

π

l

​  

 

Moment of inertia of half ring about  

=m(  

2

r  

2

 

​  

)

=m(  

2π  

2

 

l  

2

 

​  

) replacing r with  

π

l

​  

 

Moment of Inertia=  

2π  

2

 

ml  

2

 

​

Answer 2

Answer:

M$l^{2}$/2$\pi ^{2}$

Explanation:

• Length = Circumference of semicircle

• $\pi$R=l

• R=l/$\pi$

• For full circle Moment of Inertia about diameter is M$R^{2}$

• For semicircle it is M$R^{2}$/2
• Now put R=l/$\pi$

• MOI = M$l^{2}$/2$\pi$