Last Updated On 16-03-2023 - By Jeesha Sara
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
(a)
MR2
(b)
MR2
(c)
MR2
(d)
MR2
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
(a)
MR2
(b)
MR2
(c)
MR2
(d)
MR2
Question ID - 50287 :-
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
(a)
MR2
(b)
MR2
(c)
MR2
(d)
MR2
|
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
|
|||
|
(a) |
|
(b) |
|
|
(c) |
|
(d) |
|
For Ball
Using Parallel axis theorem.
Iball =
MR2+M(2R)2
=
MR2
2 Balls so 
MR2
Irod =for rod 
=
Isystem= IBall + Irod
=
MR2+
=
MR2
Next Question :
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A body is projected at t=0 with a velocity 10 ms−1 at an angle of 60° with the horizontal. The radius of its trajectory at t=1s is R. Neglecting air resistance and taking acceleration due to gravity g=10 ms−2, the value of R is: |
|||
|
(a) |
2.5 m |
(b) |
10.3 m |
|
(c) |
2.8 m |
(d) |
5.1 m |
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