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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 centers and perpendicular to their planes, will be \(\\{\mathrm{A}\\} 1: 1\) \(\\{\mathrm{B}\\} 2: 1\) \(\\{C\\} 1: 2\) \(\\{\mathrm{D}\\} 4: 1\)

Short Answer

Expert verified
The ratio of the moment of inertia for the circular rig to the circular disc is \(\{B\} 2:1\), as obtained by using the respective moment of inertia formulas \(I_{rig} = M R^2\) and \(I_{disc} = \frac{1}{2} M R^2\), and simplifying the equation for the ratio \(\frac{I_{rig}}{I_{disc}}\).

Step by step solution

01

Determine the moment of inertia formulas for the circular rig and the circular disc

We need to know the formulas for the moment of inertia for both objects. For a circular rig (hollow circle) with mass M and radius R, the moment of inertia is given by: \(I_{rig} = M R^2\). For a circular disc (solid circle) with mass M and radius R, the moment of inertia is given by: \(I_{disc} = \frac{1}{2} M R^2\).
02

Set up the equation for the ratio of moments of inertia

Now that we have the formulas for the moment of inertia, we need to set up the equation for the ratio of their moments of inertia: \(\frac{I_{rig}}{I_{disc}} = \frac{M R^2}{\frac{1}{2} M R^2}\).
03

Simplify the equation

Notice that the mass (M) and radius squared (R^2) appear in both the numerator and the denominator. We can simplify the equation by canceling them out: \(\frac{I_{rig}}{I_{disc}} = \frac{M R^2}{\frac{1}{2} M R^2} = \frac{2}{1}\). The ratio of the moment of inertia is 2:1.
04

Choose the correct answer

Now that we have simplified the equation and found the ratio of the moment of inertia for the circular rig to the circular disc, we can choose the correct answer choice: The correct answer is \(\{B\} 2:1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Disc
A circular disc is essentially a flat, solid object often used in physics to explore rotational dynamics. Think of it as a flat, filled-in circle. Understanding a circular disc is crucial because many practical objects, like wheels or gears, can be modeled as discs in theoretical physics. Calculating the moment of inertia for a circular disc is key to solving many physics problems.

The moment of inertia depends on how the mass is distributed relative to the axis of rotation. For a circular disc of mass \( M \) and radius \( R \), the formula for its moment of inertia about an axis passing through its center and perpendicular to its plane is:
  • \( I_{disc} = \frac{1}{2} M R^2 \)
This equation highlights that the mass is symmetrically distributed across a circular shape, affecting how it resists rotational motion. Remembering this formula is essential when working with rotational dynamics problems.
Circular Ring
A circular ring can be visualized as a thin band or loop with a hollow center, resembling a hoop. Unlike a solid disc, the mass of a circular ring is concentrated along its edge. This distinct mass distribution directly influences its moment of inertia.

For a circular ring with mass \( M \) and radius \( R \), the moment of inertia about an axis through its center and perpendicular to its plane is:
  • \( I_{ring} = M R^2 \)
Because all the mass is at the maximum distance from the axis, a ring has a greater moment of inertia than a disc of the same mass and radius. This quality is important when comparing rotational resistance and figuring out which objects require more force to rotate.
Physics Problems
Physics problems often require breaking down complex systems into understandable, smaller parts. When dealing with rotational dynamics, it's common to focus on objects like discs and rings because they're simpler forms that represent many real-world objects.

To solve physics problems related to rotational dynamics, you'd often:
  • Identify the object and its shape (disc, ring, etc.).
  • Determine the relevant physical quantities (mass, radius).
  • Apply the correct moment of inertia formula.
  • Simplify the equations to find the solution.
For example, knowing the formulas for moments of inertia enables you to compare how different objects will behave under rotational forces. This typically involves manipulating equations, simplifying ratios, or determining the outcomes of physical interactions in theoretical tests.
Rotational Dynamics
Rotational dynamics involves the study of objects in rotation, focusing on how forces interact with them to change their motion. It extends the principles of linear dynamics (forces and motions in a straight line) to systems rotating about an axis.

Moment of inertia plays a central role in understanding rotational dynamics. It quantifies an object's resistance to changes in its rotational velocity. The larger the moment of inertia, the harder it is to change the object’s rotation. In essence, it's the rotational equivalent of mass in linear dynamics.

When you work with problems involving rotational dynamics, you often:
  • Calculate moments of inertia for different shapes.
  • Apply Newton's second law for rotation: \( \tau = I \alpha \), where \( \tau \) is torque and \( \alpha \) is angular acceleration.
  • Analyze the energy transition between potential and kinetic energy in rotating systems.
Understanding these concepts clarifies how objects behave when subjected to rotational forces, important in fields ranging from engineering to astrophysics.

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Most popular questions from this chapter

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational \(\mathrm{K} . \mathrm{E}\). \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

A cylinder of mass \(\mathrm{M}\) has length \(\mathrm{L}\) that is 3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? \(\\{\mathrm{A}\\} 1\) \(\\{\mathrm{B}\\}(1 / \sqrt{3})\) \(\\{\mathrm{C}\\} \sqrt{3}\) \(\\{\mathrm{D}\\}(\sqrt{3} / 2)\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

In a bicycle the radius of rear wheel is twice the radius of front wheel. If \(\mathrm{r}_{\mathrm{F}}\) and \(\mathrm{r}_{\mathrm{r}}\) are the radius, \(\mathrm{v}_{\mathrm{F}}\) and \(\mathrm{v}_{\mathrm{r}}\) are speed of top most points of wheel respectively then... \(\\{\mathrm{A}\\} \mathrm{v}_{\mathrm{r}}=2 \mathrm{v}_{\mathrm{F}}\) \(\\{\mathrm{B}\\} \mathrm{v}_{\mathrm{F}}=2 \mathrm{v}_{\mathrm{r}} \quad\\{\mathrm{C}\\} \mathrm{v}_{\mathrm{F}}=\mathrm{v}_{\mathrm{r}}\) \(\\{\mathrm{D}\\} \mathrm{v}_{\mathrm{F}}>\mathrm{v}_{\mathrm{r}}\)

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