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Chapter 9: Problem 37

A particle is moving with a velocity of \(\vec{v}=(3 i+4 t j) \mathrm{m} / \mathrm{s}\) Find the ratio of tangential acceleration to that of normal acceleration at \(t=1 \mathrm{sec}\) (1) \(4 / 3\) (2) \(3 / 4\) (3) \(5 / 3\) (4) \(3 / 5\)

Short Answer

Expert verified
The ratio of tangential to normal acceleration at \(t=1\) is \(\frac{4}{3}\), which corresponds to option (1).

Step by step solution

01

Define Tangential Acceleration

The tangential acceleration is the rate of change of the magnitude of velocity. We first need to find the magnitude of the velocity: \[ |abla v| = \sqrt{3^2 + (4t)^2} = \sqrt{9 + 16t^2} \] Now, differentiate this magnitude with respect to time to get the tangential acceleration:\[ a_t = \frac{d}{dt} \left(\sqrt{9 + 16t^2}\right) = \frac{16t}{\sqrt{9 + 16t^2}} \] Substitute \(t = 1\) into this expression:\[ a_t = \frac{16 \times 1}{\sqrt{9 + 16 \times 1^2}} = \frac{16}{\sqrt{25}} = \frac{16}{5} \]
02

Define Normal Acceleration

Normal acceleration is related to the change in direction of the velocity vector. It is computed as the centripetal component, which is given by:\[ a_n = \frac{|\vec{v} \times \vec{a}|}{|abla v|} \]Where the velocity \(\vec{v}\) is \((3i + 4tj)\) and the acceleration \(\vec{a}\) is \((0i + 4j)\) after differentiating the velocity with respect to time.Compute the cross product: \[ |\vec{v} \times \vec{a}| = |(3i + 4tj) \times (0i + 4j)| = |0i \times i + 3i \times 4j + 4tj \times 4j| = 12 t \] Substitute \(t = 1\) to get:\[ |\vec{v} \times \vec{a}| = 12 \] Compute the normal acceleration:\[ a_n = \frac{12}{\sqrt{25}} = \frac{12}{5} \]
03

Compute the Ratio of Tangential to Normal Acceleration

We have the expressions for both tangential and normal accelerations at \(t=1\): \[ a_t = \frac{16}{5}, \quad a_n = \frac{12}{5} \]The ratio of tangential to normal acceleration is:\[ \frac{a_t}{a_n} = \frac{\frac{16}{5}}{\frac{12}{5}} = \frac{16}{12} = \frac{4}{3} \]
04

Choose the Correct Option

The calculated ratio of tangential to normal acceleration is \(\frac{4}{3}\).Therefore, the correct option is \((1)\) \(\frac{4}{3}\).

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Key Concepts

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

Velocity Vector
In the context of this problem, the velocity vector is key to understanding how the particle is moving. The vector can be represented as \( \vec{v} = (3i + 4tj) \) m/s. Here, \(i\) and \(j\) denote the unit vectors in the horizontal and vertical directions, respectively.
  • The \( 3i \) component tells us that there is a constant horizontal velocity of 3 m/s.
  • The \( 4t j \) component means that the vertical velocity increases with time, specifically at 4 times per each second elapsed.
Understanding the velocity vector helps in determining how the speed and direction of the particle change over time.Differences in the vector's components at various times can lead to changes in the acceleration, both tangential and normal. By examining these changes within each time period, we can move on to calculating the specific types of acceleration.
Centripetal Component
Centripetal component or normal acceleration, as used in this problem, is tied to how the direction of the velocity vector changes with time. It is given by the expression:\[ a_n = \frac{|\vec{v} \times \vec{a}|}{|\vec{v}|} \]This formula helps determine how strongly the object is pulled towards a central point due to its curved path.
  • The cross product \(|\vec{v} \times \vec{a}|\) involves finding the area of the parallelogram formed by the velocity vector \(\vec{v}\) and the acceleration vector \(\vec{a}\).
  • The magnitude of the velocity \(|\vec{v}|\) normalizes this value to ensure we know how much the object is actually being redirected.
In this scenario, understanding this concept helps calculate how fast the curve happens, which is essential for dissecting the overall motion of the particle - leading us eventually to the ratio of tangential to normal acceleration mentioned in the problem.
Cross Product
The cross product is a critical mathematical operation that helps us understand rotational properties in physics. In this problem, it is used to compute parts of the normal acceleration.
  • When we perform a cross product between the velocity vector \((3i + 4tj)\) and the acceleration vector \((0i + 4j)\), it provides a measure of how perpendicular the vectors are to one another.
  • This calculation yields \(12t\), showing how rotational effects might impact the motion.
Since the cross product results in a vector perpendicular to both initial vectors, it provides insight into how the motion operates around a circular path, shaping how we compute normal acceleration.
Magnitude of Velocity
The magnitude of velocity provides the raw speed of an object, disregarding its direction. Calculating it is essential for both tangential and normal acceleration.
  • For the given velocity vector \(\vec{v} = (3i + 4tj)\), its magnitude is found using the formula \(|\vec{v}| = \sqrt{3^2 + (4t)^2}\), simplifying to \(\sqrt{9 + 16t^2}\).
  • This magnitude tells us the detailed breakdown of speed stemming from both horizontal and vertical motions.
Calculating the magnitude helps in understanding how fast the object is moving overall, which remains a constant point of reference throughout the particle's trajectory and consistent for calculating accelerations in physics.

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Most popular questions from this chapter

Two identical particics we string which passes through a hole at the center of a table of the particles is made to move in a circle on that string which passes througit One of the particles is made to move in a circle on the tabl with angular velocity \(\omega\), and the other is made to \(\mathrm{m}\), in a horizontal circle as a contact pendulum with angular velocity pendulum with angular velocity \(\omega_{2} .\) If \(l_{1}\) and \(l_{2}\) are the length of he string over and under the able, then in order that particle under the table neither moves own nor moves up, the ratio \(l_{1} / l_{2}\) is (1) \(\frac{\omega_{1}}{\omega_{2}}\) (2) \(\frac{\omega_{2}}{\omega_{1}}\) (3) \(\frac{\omega_{1}^{2}}{\omega_{2}^{2}}\) (4) \(\frac{\omega_{2}^{2}}{\omega_{1}^{2}}\)

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A stone is thrown from point \(O\) on ground with velocity \(\vec{v}=5 \hat{i}+10 \hat{j} \mathrm{~m} / \mathrm{s}\), where \(x\) is horizontal and \(y\) is vertical. Its. angular velocity about \(O\) when the stone is at maximum height of its trajectory, is (i) \(\frac{1}{2} \mathrm{rad} / \mathrm{s}\) (2) \(\frac{1}{\sqrt{2}} \mathrm{rad} / \mathrm{s}\) (3) \(1 \mathrm{rad} / \mathrm{s}\) (4) \(\sqrt{2} \mathrm{rad} / \mathrm{s}\)
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