Problem 562
Question
A ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is melted and then molded in to a sphere then the moment of inertia of the sphere will be..... \(\\{\mathrm{A}\\}\) more than that of the ring \\{B \\} Less than that of the ring \(\\{\mathrm{C}\\}\) Equal to that of the ring \\{D\\} None of these
Step-by-Step Solution
Verified Answer
The moment of inertia of the sphere is less than that of the ring. The correct answer is (B) Less than that of the ring.
1Step 1: Calculate the moment of inertia of the ring
The moment of inertia of a ring with mass \(M\) and radius \(r\) is given by the formula:
\(I_\text{ring} = Mr^2\)
In our case, we have the mass (\(M\)) and radius (\(r\)) of the ring, so we can directly apply this formula:
\(I_\text{ring} = M r^2\)
2Step 2: Calculate the moment of inertia of the sphere
When the ring is melted and molded into a sphere, its moment of inertia will be given by the formula:
\(I_\text{sphere} = \frac{2}{5}M R^2\)
Since the mass and radius of the sphere are both equal to the mass and radius of the ring, we can substitute those values:
\(I_\text{sphere} = \frac{2}{5}M r^2\)
3Step 3: Compare the moments of inertia
Now we have both moments of inertia and can compare them:
\(I_\text{ring} = M r^2\) and \(I_\text{sphere} = \frac{2}{5}M r^2\)
Start by dividing both sides by \(M r^2\):
\(\frac{I_\text{ring}}{M r^2} = 1\)
and
\(\frac{I_\text{sphere}}{M r^2} = \frac{2}{5}\)
Now we compare these two ratios, noticing that
\(\frac{2}{5} < 1\)
This means that
\(I_\text{sphere} = \frac{2}{5}M r^2 < M r^2 = I_\text{ring}\)
Hence, the moment of inertia of the sphere is less than that of the ring.
4Step 4: Answer
The correct answer is (B) Less than that of the ring.
Key Concepts
Ring and Sphere TransformationPhysics JEE ProblemRotational Motion Concepts
Ring and Sphere Transformation
When discussing the transformation from a ring to a sphere, it is important to understand the concept of redistributing mass. Initially, we have a ring, which is essentially a thin, circular band of a specified mass \( M \) and radius \( r \).
- The mass of the ring is uniformly distributed along its circumference.
- The moment of inertia of the ring is determined by how far its mass is from its rotational axis, which is given by the formula \( I_\text{ring} = Mr^2 \).
Physics JEE Problem
The JEE (Joint Entrance Examination) is a highly standardized test in India, which includes challenging problems in physics to test a student's understanding of various concepts. One type of problem you might encounter involves moments of inertia, such as the one given in the exercise: a ring that is molded into a sphere.
- Such problems test your ability to understand and manipulate formulas for moment of inertia.
- They also require you to apply theoretical physics concepts to practical changes in shape and mass distribution.
- Knowing how to derive and compare different inertia values is crucial for solving these problems correctly.
Rotational Motion Concepts
Rotational motion is a fundamental topic in mechanics within physics. It extends the idea of linear motion to objects that rotate around an axis, meaning concepts like inertia adapt to consider distributed mass about a rotational center.Understanding rotational motion begins with the concept of the moment of inertia, a measure of an object's resistance to changes in its rotational state.
- The moment of inertia depends not just on mass but also on the shape and axis of rotation.
- For simple geometric shapes, specific formulas calculate the moment of inertia, such as \( Mr^2 \) for a ring and \( \frac{2}{5}MR^2 \) for a sphere.
Other exercises in this chapter
Problem 560
One circular rig and one circular disc both are having the same mass and radius. The ratio of their moment of inertia about the axes passing through their cente
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One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters
View solution Problem 563
A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to
View solution Problem 564
One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendi
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