/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 149 The particle of mass \(m=1.2 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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The particle of mass \(m=1.2 \mathrm{kg}\) is attached to the end of the light rigid bar of length \(L=0.6 \mathrm{m} .\) The system is released from rest while in the horizontal position shown, at which the torsional spring is undeflected. The bar is then observed to rotate \(30^{\circ}\) before stopping momentarily. \((a)\) Determine the value of the torsional spring constant \(k_{T} .(b)\) For this value of \(k_{T},\) determine the speed \(v\) of the particle when \(\theta=15^{\circ}\).

Short Answer

Expert verified
(a) Calculate the torsional spring constant using conservation of energy. (b) Calculate the particle's speed using the stored spring energy at \( \theta=15^{\circ} \).

Step by step solution

01

Identify Given Variables

The mass of the particle is \( m = 1.2 \, \text{kg} \), the length of the bar is \( L = 0.6 \, \text{m} \), and the particle rotates through \( \theta = 30^{\circ} \) before stopping. The initial angle is \( \theta_0 = 0^{\circ} \), and the final angle at rest is \( \theta_f = 30^{\circ} \). The torsional spring constant is \( k_T \), which we need to determine.
02

Apply Energy Conservation

Using energy conservation, we know that the initial potential energy equals the sum of the kinetic energy and the potential energy stored in the spring at \( \theta = 30^{\circ} \). Initial potential energy is zero since the system starts from rest. So,\[0 = -mgL(1 - \cos \theta_f) + \frac{1}{2}k_T \theta_f^2.\]
03

Solve for Torsional Spring Constant \( k_T \)

We substitute values into the energy equation from Step 2: \( g = 9.8 \, \text{m/s}^2 \), \( \theta_f = 30^{\circ} = \frac{\pi}{6} \).\[-(1.2)(9.8)(0.6)(1 - \cos(\frac{\pi}{6})) = \frac{1}{2}k_T(\frac{\pi}{6})^2.\] Simplify and solve for \( k_T \):\[ k_T = \frac{2 \times 1.2 \times 9.8 \times 0.6 \times (1 - \cos(30^{\circ}))}{(\frac{\pi}{6})^2}.\]
04

Determine Effective Energy at \( \theta=15^{\circ} \)

When \( \theta = 15^{\circ} \), apply the energy equation:\[-mgL(1 - \cos \theta) + \frac{1}{2}k_T \theta^2 + \frac{1}{2}mv^2 = 0.\] Substituting previous calculated \( k_T \), solve for velocity \( v \):\[-(1.2)(9.8)(0.6)(1 - \cos(15^{\circ})) + \frac{1}{2}k_T (\frac{\pi}{12})^2 + \frac{1}{2}(1.2)v^2 = 0.\] Rearrange and solve for \( v \).

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle stating that energy within a closed system remains constant over time. It reminds us that energy cannot be created or destroyed but only transformed or transferred. In our problem, energy conservation helps us track how energy moves between potential, kinetic, and elastic (spring) energy forms.

Initially, the energy in the system is purely gravitational potential energy, as the particle is at rest and the torsional spring is undeflected. As the bar rotates, this potential energy is converted into the elastic energy of the torsional spring, and eventually, some kinetic energy when the system moves to an angle of \( \theta = 15^{\circ} \).

Here’s how we do it:
  • Initial energy is zero (since the system starts from rest).
  • At \( \theta = 30^{\circ}, \) the potential energy due to gravity has decreased and is now stored as elastic potential energy in the spring.
  • Using the formula \(-mgL(1 - \cos(\theta)) + \frac{1}{2}k_T \theta^2 \), we equate gravitational potential energy with spring energy to find the torsional spring constant \(k_T\).
  • Later, at \( \theta = 15^{\circ}, \) we calculate the kinetic energy to determine the speed \(v\) of the particle.
This physical conservation guides the flowing energy throughout the system, allowing us to solve complex dynamics with straightforward principles.
Rotational Dynamics
Rotational Dynamics involves the study of objects rotating around a pivot point and the forces and torques that influence such motion. This concept is crucial in understanding everything from spinning wheels to orbiting satellites. Here, rotational dynamics allows us to understand how the particle attached to a rotating bar behaves.

Key elements include:
  • Angular Motion: This is similar to linear motion but in a rotational sense. Our particle moves through specific angles, measured in degrees or radians.
  • Torque: This is the rotational equivalent of force, often exerted by springs or other factors to cause rotation. For instance, the torsional spring provides torque that affects the bar's rotation.
  • Moment of Inertia: This is a measure of an object's resistance to changes in its rotational motion, analogous to mass in linear motion.
In the problem, the rotational dynamics are mapped out as the particle moves from 0 to \(30^{\circ}\), and then to \(15^{\circ}\). We consider:
  • Gravitational torque generating force as the particle descends.
  • The restoring torque from the torsional spring opposing motion.
By evaluating these dynamics, we comprehend how energy shifts and how the bar continues through space.
Angular Displacement
Angular displacement measures how much an object rotates over an angle. It's a more intuitive term than linear displacement, as it calculates rotation typically using degrees or radians.

In our example, angular displacement tells us that the bar rotates \(\theta = 30^{\circ}\) before momentarily stopping, marking the change in position from its starting angle (\(\theta_0 = 0^{\circ}\)) to its rest position (\(\theta_f = 30^{\circ}\)).

Understanding angular displacement involves:
  • Measurement in Radians: Although angles can be described in degrees, radians are more commonly used in calculations due to their natural relationship with circular motion. For instance, \(30^{\circ}\) equals \(\frac{\pi}{6}\) radians.
  • Relation to Energy: In our task, as angular displacement occurs, energy transitions, helped by torque and the torsional spring, which stores potential energy based on its angular deflection. Increasing displacement leads to increased stored energy.
  • Units and Conversion: Mastering conversions between degrees and radians for efficient computation is essential for resolving dynamic equations.
Angular displacement serves as the cornerstone to connecting energy states within this rotating system, through both conceptual understanding and practical calculation.

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