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The moment of inertia of a thin uniform circular disc about one of the diameters is I. Its moment of inertia about an axis perpendicular to the circular surface and passing through the centre is
- a)√2I
- b)2I
- c)I ∕ √ 2
- d)I ∕ 2
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The moment of inertia of a thin uniform circular disc about one of the...
I/2
b) I
c) 2I
d) 4I
The answer is b) I.
Explanation:
The moment of inertia of a thin uniform circular disc about one of the diameters is given by the formula:
I = (1/4)MR²
where M is the mass of the disc and R is its radius.
The moment of inertia about an axis perpendicular to the circular surface and passing through the centre can be found using the parallel axis theorem:
I' = I + Md²
where d is the distance between the two axes.
In this case, d is equal to the radius of the disc, since the axis of rotation passes through the centre. Therefore, we have:
I' = I + MR²
Substituting the value of I from the given formula, we get:
I' = (1/4)MR² + MR²
Simplifying, we get:
I' = (5/4)MR²
Since M and R are constants for the disc, we can write:
I' = kI
where k is a constant. Therefore, the moment of inertia about an axis perpendicular to the circular surface and passing through the centre is equal to the moment of inertia about one of the diameters, which is I.
b) I
c) 2I
d) 4I
The answer is b) I.
Explanation:
The moment of inertia of a thin uniform circular disc about one of the diameters is given by the formula:
I = (1/4)MR²
where M is the mass of the disc and R is its radius.
The moment of inertia about an axis perpendicular to the circular surface and passing through the centre can be found using the parallel axis theorem:
I' = I + Md²
where d is the distance between the two axes.
In this case, d is equal to the radius of the disc, since the axis of rotation passes through the centre. Therefore, we have:
I' = I + MR²
Substituting the value of I from the given formula, we get:
I' = (1/4)MR² + MR²
Simplifying, we get:
I' = (5/4)MR²
Since M and R are constants for the disc, we can write:
I' = kI
where k is a constant. Therefore, the moment of inertia about an axis perpendicular to the circular surface and passing through the centre is equal to the moment of inertia about one of the diameters, which is I.
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Community Answer
The moment of inertia of a thin uniform circular disc about one of the...
I/2
b) I
c) 2I
d) 3I/2
The correct answer is: b) I
Explanation:
The moment of inertia of a thin uniform circular disc about one of its diameters is given by:
I = (1/4)MR^2
where M is the mass of the disc and R is the radius.
The moment of inertia of the disc about an axis perpendicular to the circular surface and passing through the centre can be found using the parallel axis theorem:
I' = I + Md^2
where d is the distance between the two axes.
Since the axis passing through the centre is perpendicular to the diameter, the distance d is equal to R/2.
Substituting the values, we get:
I' = I + M(R/2)^2
I' = I + (1/4)MR^2
I' = (1/4)MR^2 + (1/4)MR^2
I' = (1/2)MR^2
Since M and R are constants, the moment of inertia about an axis perpendicular to the circular surface and passing through the centre is equal to:
I' = (1/2)MR^2 = I
Therefore, the correct answer is b) I.
b) I
c) 2I
d) 3I/2
The correct answer is: b) I
Explanation:
The moment of inertia of a thin uniform circular disc about one of its diameters is given by:
I = (1/4)MR^2
where M is the mass of the disc and R is the radius.
The moment of inertia of the disc about an axis perpendicular to the circular surface and passing through the centre can be found using the parallel axis theorem:
I' = I + Md^2
where d is the distance between the two axes.
Since the axis passing through the centre is perpendicular to the diameter, the distance d is equal to R/2.
Substituting the values, we get:
I' = I + M(R/2)^2
I' = I + (1/4)MR^2
I' = (1/4)MR^2 + (1/4)MR^2
I' = (1/2)MR^2
Since M and R are constants, the moment of inertia about an axis perpendicular to the circular surface and passing through the centre is equal to:
I' = (1/2)MR^2 = I
Therefore, the correct answer is b) I.
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The moment of inertia of a thin uniform circular disc about one of the diameters is I. Its moment of inertia about an axis perpendicular to the circular surface and passing through the centre isa)√2Ib)2Ic)I √ 2d)I 2Correct answer is option 'B'. Can you explain this answer?
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