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Chapter 3: Problem 75

A rock tied to a rope moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R \cos \omega t \quad y(t)=R \sin \omega t $$ where \(R\) and \(\omega\) are constants. (a) Show that the rock's distance from the origin is constant and equal to \(R-\) that is, that its path is a circle of radius \(R .(b)\) Show that at every point the rock's velocity is perpendicular to its position vector. (c) Show that the rock's acceleration is always opposite in direction to its position vector and has magnitude \(\omega^{2} R .\) (d) Show that the magnitude of the rock's velocity is constant and equal to \(\omega R\) . (e) Combine the results of parts \((\mathrm{c})\) and \((\mathrm{d})\) to show that the rock's acceleration has constant magnitude \(v^{2} / R .\)

Short Answer

Expert verified
The rock's path is a circle, its velocity is perpendicular to position, acceleration is opposite to position, and both velocity and a calculated acceleration magnitude are constant.

Step by step solution

01

Calculate the Distance from the Origin

To prove the rock's distance from the origin is constant, we'll calculate the distance using the Pythagorean theorem. The position vector is \((x(t), y(t)) = (R \cos \omega t, R \sin \omega t)\). The distance from the origin is the magnitude of this vector:\[d(t) = \sqrt{x(t)^2 + y(t)^2} \= \sqrt{(R \cos \omega t)^2 + (R \sin \omega t)^2} \= \sqrt{R^2 \cos^2 \omega t + R^2 \sin^2 \omega t} \= \sqrt{R^2 (\cos^2 \omega t + \sin^2 \omega t)} = \sqrt{R^2} = R.\] Thus, the distance is always \(R\), which confirms the path is a circle of radius \(R\).
02

Show Velocity is Perpendicular to Position

The velocity vector \(\vec{v}(t)\) is the derivative of the position vector with respect to time. Differentiating the position components gives \[v_x(t) = \frac{d}{dt}(R \cos \omega t) = -R \omega \sin \omega t\] \[v_y(t) = \frac{d}{dt}(R \sin \omega t) = R \omega \cos \omega t.\] Thus, \(\vec{v}(t) = (-R \omega \sin \omega t, R \omega \cos \omega t)\). The dot product \(\vec{v}(t) \cdot \vec{r}(t)\) is \((-R \omega \sin \omega t \cdot R \cos \omega t) + (R \omega \cos \omega t \cdot R \sin \omega t) = 0\). The dot product being zero means the vectors are perpendicular.
03

Show Acceleration is Opposite to Position

Acceleration \(\vec{a}(t)\) is the derivative of the velocity vector. Differentiating \(\vec{v}(t)\), we have \(a_x(t) = \frac{d}{dt}(-R \omega \sin \omega t) = -R \omega^2 \cos \omega t\) and \(a_y(t) = \frac{d}{dt}(R \omega \cos \omega t) = -R \omega^2 \sin \omega t\). Therefore, \(\vec{a}(t) = (-R \omega^2 \cos \omega t, -R \omega^2 \sin \omega t)\), which is \(-\omega^2 (x(t), y(t))\). Hence, it is opposite in direction with magnitude \(\omega^2 R\).
04

Show Constant Velocity Magnitude

The magnitude of the velocity vector \(\vec{v}(t)\) is \[v(t) = \sqrt{(-R \omega \sin \omega t)^2 + (R \omega \cos \omega t)^2}\] \[= \sqrt{R^2 \omega^2 \sin^2 \omega t + R^2 \omega^2 \cos^2 \omega t}\] \[= \omega R \sqrt{\sin^2 \omega t + \cos^2 \omega t} = \omega R.\] Therefore, the velocity magnitude is constant and equals \(\omega R\).
05

Use Velocity to Confirm Acceleration Magnitude

Using the results from steps 3 and 4, the magnitude of the acceleration from step 3 is \(\omega^2 R\). Let \(v = \omega R\). Then substituting \(v^2 = (\omega R)^2 = \omega^2 R^2\), results in the acceleration magnitude as \(\frac{v^2}{R} = \frac{\omega^2 R^2}{R} = \omega^2 R\), confirming the result.

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Key Concepts

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

Velocity Perpendicular to Position
In circular motion, understanding the relationship between velocity and position is crucial. When an object, like a rock tied to a rope, follows a circular path, its velocity is always tangent to the circle at any given point. This tangent nature ensures that the velocity vector is perpendicular to the position vector, centered at the origin of the circle.

The position vector, represented as \( \vec{r}(t) = (x(t), y(t)) \), points from the center of the circle to the object's location. As derived, the velocity vector \( \vec{v}(t) = (-R \omega \sin \omega t, R \omega \cos \omega t) \), results from differentiating the position over time.

The perpendicularity is proven through their dot product being zero:
  • \( \vec{v}(t) \cdot \vec{r}(t) = (-R \omega \sin \omega t \cdot R \cos \omega t) + (R \omega \cos \omega t \cdot R \sin \omega t) = 0 \).
This zero result means they are perpendicular, explaining why the rock moves in a circle without straying from its path.
Acceleration Opposite to Position
Acceleration in circular motion is a concept that often turns the direction of thinking. Here, the acceleration is not in the direction of the motion but rather opposite to the position vector. This means as the rock moves, its acceleration points inward, toward the center of the circle.

Calculating the acceleration vector, we differentiate the velocity vector:\( \vec{a}(t) = (-R \omega^2 \cos \omega t, -R \omega^2 \sin \omega t) \). These components show it as \( -\omega^2 (x(t), y(t)) \). The negative sign indicates it is in the opposite direction to the position vector \( (x(t), y(t)) \).

  • The magnitude of the acceleration is \( \omega^2 R \), which balances the inward pull keeping the rock in its circular path.
This opposing nature is vital for maintaining the constant curved trajectory.
Constant Velocity Magnitude
One fascinating characteristic of circular motion is the constant magnitude of velocity, even as direction changes swiftly. Using the previously calculated velocity vector, its magnitude is given by:

\[ v(t) = \sqrt{(-R \omega \sin \omega t)^2 + (R \omega \cos \omega t)^2} \]

Simplifies to \( \omega R \), a remarkable result proving the velocity does not vary in size. This consistency in velocity magnitude, despite the continuous directional change, is what allows the rock to trace out a perfect circle.

  • It embodies the principle that speed remains steady while direction alters.
Therefore, it revolves in uniform circular motion, essential for understanding phenomena like orbits and rotations.
Pythagorean Theorem in Physics
The foundational Pythagorean theorem is not just limited to geometry but extends its application to physics, especially in circular motion. It provides the essential tool for determining distances, which are instrumental in analyzing circular paths.

When assessing the rock tied to a rope, the position components form a right triangle with the origin, allowing the use of the Pythagorean theorem:

\[ d(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{(R \cos \omega t)^2 + (R \sin \omega t)^2} = R \].

This approach demonstrates how the distance from the center is constant, solidifying that the rock moves in a circle with radius \( R \).

  • It assures us that even as variables shift due to time, certain relationships remain unaffected, like the radial distance.
This concept is ubiquitous, aiding various physics problems involving periodic motion and vector analysis.

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Most popular questions from this chapter

For no apparent reason, a poodle is running at a constant speed of \(v=5.00 \mathrm{m} / \mathrm{s}\) in a circle with radius \(R=2.50 \mathrm{m} .\) Let \(\vec{v}_{1}\) be the velocity vector at time \(t_{1},\) and let \(\vec{v}_{2}\) be the velocity vector at time \(t_{2}\) . Consider \(\Delta \vec{v}=\vec{v}_{2}-\vec{v}_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Recall that \(\vec{a}_{\mathrm{av}}=\Delta \vec{v} / \Delta t .\) For \(\Delta t=0.5 \mathrm{s}, 0.1 \mathrm{s},\) and 0.05 \(\mathrm{s}\) , calculate the magnitude (to four significant figures) and direction (relative to \(\vec{v}_{1} )\) of the average acceleration \(\vec{d}_{\mathrm{ev}}\) . Compare your results to the general expression for the instantaneous acceleration \(\vec{a}\) for uniform circular motion that is derived in the text.

Dynamite! A demolition crew uses dynamite to blow an old building apart. Debris from the explosion fies off in all directions and is later found at distances as far as 50 \(\mathrm{m}\) from the explosion. Estimate the maximum speed at which debris was blown outward by the explosion. Describe any assumptions that you make.

Raindrops. When a train's velocity is 120 \(\mathrm{m} / \mathrm{s}\) eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined \(30.0^{\circ}\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

A man stands on the roof of a 15.0 -m-tall building and throws a rock with a velocity of magnitude 30.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock; \((b)\) the magnitude of the velocity of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t\) \(v_{x}-t_{2}\) and \(v_{y}-t\) graphs for the motion.
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