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Chapter 10: Problem 6

As \(T=2 \pi \sqrt{\frac{I}{c}}\) where \(I=\) M.I. of the system, \(c=\) torsional constant \(\therefore \quad 2=2 \pi \sqrt{\frac{0.2}{c}} \Rightarrow \frac{2 \pi}{\sqrt{c}}=\frac{2}{\sqrt{0.2}}\) Now when new disc is loaded with another disc, then \(T=2 \cdot 5=2 \pi \sqrt{\frac{I^{\prime}}{c}}=\frac{2 \pi}{\sqrt{c}} \sqrt{l^{\prime}}\) \(\Rightarrow \quad 2.5=\frac{2}{\sqrt{0.2}}\), from (1) \(\Rightarrow \quad 6.25=\frac{4}{0.2} \times l^{\prime} \quad \therefore \quad l^{\prime}=0.315\) Now required moment of inertia \(=0.315-0.2=0.115 \mathrm{~kg}-\mathrm{m}^{2}\)

Short Answer

Expert verified
The required moment of inertia is 0.115 kg-m虏.

Step by step solution

01

Write down the given equation

The original equation for the period of torsional oscillation is given by:\[ T = 2 \pi \sqrt{\frac{I}{c}} \]where \(I\) is the moment of inertia and \(c\) is the torsional constant.
02

Rearrange the initial condition

We are given that:\[ 2 = 2 \pi \sqrt{\frac{0.2}{c}} \]This can be rearranged as:\[ \frac{2 \pi}{\sqrt{c}} = \frac{2}{\sqrt{0.2}} \]
03

Solve for the modified condition with new disc

When a new disc is added, the new period \(T\) becomes 2.5:\[ T = 2.5= 2 \pi \sqrt{\frac{I'}{c}} \]We know from Step 2 that:\[ \frac{2 \pi}{\sqrt{c}} = \frac{2}{\sqrt{0.2}} \]Thus,\[ 2.5 = \frac{2}{\sqrt{0.2}} \sqrt{I'} \]
04

Solve for \(I'\)

Square both sides to eliminate the square root:\[ (2.5)^2 = \left( \frac{2}{\sqrt{0.2}} \right)^2 \times I' \]This simplifies to:\[ 6.25 = \frac{4}{0.2} \times I' \]
05

Final calculation for \(I'\)

Solving for \(I'\), we get:\[ I' = \frac{6.25 \times 0.2}{4} = 0.315 \] kg-m虏
06

Calculate the required moment of inertia

The required moment of inertia, when the new disc is loaded, is:\[ I' - 0.2 = 0.315 - 0.2 = 0.115 \] kg-m虏

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Key Concepts

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

Moment of Inertia
The moment of inertia, often abbreviated as M.I., is a critical concept in mechanics. Essentially, it measures how difficult it is to change an object's rotational motion. Think of it like mass in linear motion, where larger mass means more inertia. For rotational scenarios, objects with higher moments of inertia require more torque to achieve the same angular acceleration as objects with lower moments of inertia. This is why the same force might rotate a small pencil easily, but would barely move a large book.

In terms of formulas, the moment of inertia is represented by the symbol \(I\), and is intrinsic to calculating the period of torsional oscillations, as seen in the equation: \[T = 2 \pi \sqrt{\frac{I}{c}} \] Here, the symbol \(I\) is pivotal for determining how the system will behave under oscillation. Identifying the moment of inertia involves knowing the mass distribution relative to the axis of rotation, and it varies depending on the shape and configuration of the object.
Torsional Constant
The torsional constant, denoted as \(c\), plays a crucial role in understanding torsional oscillation systems. This constant measures the stiffness of the spring or wire in a system, much like a spring constant in a linear motion context. A higher torsional constant means the material is stiffer and resists twisting more rigorously.

When you see an equation like \[T = 2 \pi \sqrt{\frac{I}{c}}\] understanding \(c\) is vital because it balances out the relationship between the moment of inertia and the period of oscillation. If you increase \(c\) while keeping \(I\) constant, the period \(T\) of oscillation decreases, implying the system oscillates more quickly.

Grasping the importance of \(c\) helps in designing systems that need specific oscillatory behaviors, such as measuring instruments and machinery requiring stability.
Period of Oscillation
The period of oscillation, represented by \(T\), is the time taken for one complete cycle of an oscillating system. This concept is crucial for engineers and scientists because it helps predict system behavior over time. In our given formula \[T = 2 \pi \sqrt{\frac{I}{c}}\] \(T\) depends on both the moment of inertia \(I\) and the torsional constant \(c\).

A longer period means the system takes more time to complete one oscillation, which can be indicative of a large moment of inertia or a low torsional constant. Conversely, a shorter period implies the opposite.

Understanding the period is essential, especially when precision timing is needed in devices. For example, clocks rely on well-calculated periods of oscillation to keep accurate time. Thus, manipulating \(I\) and \(c\) to alter \(T\) can finely tune systems to fit specific requirements or constraints.

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Most popular questions from this chapter

Statement-1 : In a simple harmonic oscillation the average potential encrgy in one time period is cqual to the average kinetic energy in this period. Statement-2 : The average kinetic energy for one oscillation is equal to half of total mechanical energy.

In the state of rotation of the system as a whole there are two forces which act on the sphere: (i) elastic and (ii) centripetal. The equation of motion of the sphere is \(\frac{m d^{2} x}{d t^{2}}=-k x+m \omega^{2} x ;\) (here \(x\) sunds for displacement) or \(\frac{d^{2} x}{d t^{2}}=-\left(\frac{k}{m}-\omega^{2}\right) x\) i.e., the sphere will execute S.H.M. with $$ \begin{aligned} \omega_{n} &=\sqrt{\frac{k}{m}-\omega^{2}} \\ \therefore \quad T &=\frac{2 \pi}{\omega_{o}}=\frac{2 \pi}{\sqrt{k / m-\omega^{2}}} \\ &=\frac{2 \times 3.14}{\sqrt{\frac{20}{0.2}-(4.4 \times 4.4)}} \approx 0.7 \mathrm{~s} \end{aligned} $$ The oscillation will stop when \(T=\infty\), This is posible when $$ \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{20}{0.2}}=10 \mathrm{rad} / \mathrm{s} $$

\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) Now, According to question, \(v=v_{o}=220 \mathrm{II} z\) \(\therefore \frac{v}{2 L}=220 \mathrm{~Hz} \quad[\quad L=40 \mathrm{~cm}\) \(m=3.2 \mathrm{gm}\) \(\sqrt{\frac{T}{\delta}} \times \frac{1}{2 L}=220 \mathrm{II} z\) 1. \(\sqrt{\frac{T}{\rho}}=220 \times 2 L\) \(\therefore \quad \sqrt{T}=\sqrt{\rho} \times 220 \times 2 L$$$ \begin{aligned} &\therefore \quad T=\rho \times(220 \times 2 L)^{2}=\frac{4}{5} \times 10^{-2}(22 \times 2 \times 4)^{2} \\ &=\frac{4}{5} \times 10^{-2} \times 30976=2478.8 \times 10^{-2} \\ &\text { Now suress }=\frac{F}{\Lambda}=\frac{T}{\Lambda}=\frac{2478.8 \times 10^{-2}}{\left(10^{-3}\right)^{2}} \\ &=2478.8 \times 10^{4}=\mathrm{N} / \mathrm{m}^{2} \end{aligned} $$ and strain $$ =\frac{0.05}{40} \times \frac{10^{2}}{10^{2}}=\frac{5}{40} \times 10^{-2}=\frac{1}{8} \times 10^{-2} $$ \)\therefore$ (Youngs modulus) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$

Statement-1: If force acting on a particle is given by \(F=-k x^{2}\), then the motion of the particle is periodic but not the simple harmonic oscillation. Statement-2 : For the simple harmonic oscillation the force must be proportional to the displacement of the particle from the mean position.

Three sources of sound produce sound of the same intensity but their frequencies are 400,401 , \(402 \mathrm{~Hz}\), respectively. How many beats are heard per second?
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