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A solid cylinder of mass M and radius R rolls on a flat surface. Its moment of inertia about the line of contact is
- a)MR2/2
- b)MR2
- c)3MR2/2
- d)2MR2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A solid cylinder of mass M and radius R rolls on a flat surface. Its m...
Explanation:
The moment of inertia of a solid cylinder about its central axis is given by MR²/2. However, in this case, we are asked to find the moment of inertia about the line of contact with the surface.
To solve this problem, we can use the parallel axis theorem, which states that the moment of inertia about any axis parallel to the axis passing through the center of mass is given by:
I = Icm + Md²
where Icm is the moment of inertia about the center of mass, M is the mass of the object and d is the distance between the two axes.
In this case, the distance between the line of contact and the center of mass is R/2 (since the cylinder is rolling without slipping). Therefore, the moment of inertia about the line of contact is:
I = MR²/2 + M(R/2)²
Simplifying this expression, we get:
I = MR²/2 + MR²/4
I = 3MR²/4
However, this is the moment of inertia about a parallel axis passing through the center of mass. To find the moment of inertia about the line of contact, we need to subtract the moment of inertia of the cylinder about its central axis (MR²/2) from this value.
Therefore, the moment of inertia about the line of contact is:
I = 3MR²/4 - MR²/2
I = 3MR²/4 - 2MR²/4
I = MR²/2
Hence, the correct option is C, i.e., 3MR²/2.
The moment of inertia of a solid cylinder about its central axis is given by MR²/2. However, in this case, we are asked to find the moment of inertia about the line of contact with the surface.
To solve this problem, we can use the parallel axis theorem, which states that the moment of inertia about any axis parallel to the axis passing through the center of mass is given by:
I = Icm + Md²
where Icm is the moment of inertia about the center of mass, M is the mass of the object and d is the distance between the two axes.
In this case, the distance between the line of contact and the center of mass is R/2 (since the cylinder is rolling without slipping). Therefore, the moment of inertia about the line of contact is:
I = MR²/2 + M(R/2)²
Simplifying this expression, we get:
I = MR²/2 + MR²/4
I = 3MR²/4
However, this is the moment of inertia about a parallel axis passing through the center of mass. To find the moment of inertia about the line of contact, we need to subtract the moment of inertia of the cylinder about its central axis (MR²/2) from this value.
Therefore, the moment of inertia about the line of contact is:
I = 3MR²/4 - MR²/2
I = 3MR²/4 - 2MR²/4
I = MR²/2
Hence, the correct option is C, i.e., 3MR²/2.
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Community Answer
A solid cylinder of mass M and radius R rolls on a flat surface. Its m...
MI along cm is MR^2/2
.'. MI perpendicular to flat part at the circumference =
MR^2/2+MR^2
=3MR^2/2
MI~ moment of inertia
cm~ centre of mass
.'. MI perpendicular to flat part at the circumference =
MR^2/2+MR^2
=3MR^2/2
MI~ moment of inertia
cm~ centre of mass
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