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Chapter 4: Problem 68

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1}\), which is less than one period?

Short Answer

Expert verified
(a) Centripetal acceleration is approximately 5.24 m/s虏. (b) Average acceleration is (-2.00 i - 2.67 j) m/s虏.

Step by step solution

01

Calculate the magnitude of initial and final velocities

The magnitude of a vector \( \vec{v} = a \hat{i} + b \hat{j} \) is given by \( \sqrt{a^2 + b^2} \). For \( \vec{v}_1 = (3.00 \mathrm{~m/s}) \hat{\mathrm{i}} + (4.00 \mathrm{~m/s}) \hat{\mathrm{j}} \), its magnitude is: \[ |\vec{v}_1| = \sqrt{3.00^2 + 4.00^2} = \sqrt{9 + 16} = 5.00 \mathrm{~m/s} \] Similarly, for \( \vec{v}_2 = (-3.00 \mathrm{~m/s}) \hat{\mathrm{i}} + (-4.00 \mathrm{~m/s}) \hat{\mathrm{j}} \), its magnitude is: \[ |\vec{v}_2| = \sqrt{(-3.00)^2 + (-4.00)^2} = \sqrt{9 + 16} = 5.00 \mathrm{~m/s} \].
02

Determine the formula for centripetal acceleration

Centripetal acceleration \( a_c \) for an object in circular motion is given by \[ a_c = \frac{v^2}{r} \] where \( v \) is the speed and \( r \) is the radius of the circular path.
03

Use uniform circular motion property

Since it's uniform circular motion, the speed is constant. The velocities \( \vec{v}_1 \) and \( \vec{v}_2 \) have the same magnitude, confirming uniform circular motion.
04

Find the angle between \( \vec{v}_1 \) and \( \vec{v}_2 \)

Since \( \vec{v}_1 \) and \( \vec{v}_2 \) are 180 degrees apart (opposite directions), the angle \( \Delta \theta = \pi \) radians, indicating half a circle or half a period of motion. This simplifies our calculations.
05

Calculate centripetal acceleration

Using the relationship between \( v \), \( r \), and the fact that the change in angle \( \Delta \theta = \pi \) over 3 seconds: \( T = 2(t_2 - t_1) = 2(5-2) = 6 \) seconds for a full period. Therefore: \[ r = \frac{|\vec{v}|T}{2\pi} = \frac{5 \times 6}{2\pi} \approx 4.77 \mathrm{~m} \] Now, compute \( a_c \): \[ a_c = \frac{v^2}{r} = \frac{5^2}{4.77} \approx 5.24 \mathrm{~m/s^2} \].
06

Calculate average acceleration

The average acceleration \( \vec{a}_{avg} \) is given by the change in velocity over the time interval: \[ \vec{a}_{avg} = \frac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1} = \frac{(-3.00\hat{i} - 3.00\hat{i}) + (-4.00\hat{j} - 4.00\hat{j})}{5.00 - 2.00} \] \[ \vec{a}_{avg} = \frac{(-6.00 \hat{i} - 8.00 \hat{j})}{3.00} = (-2.00 \hat{i} - 2.67 \hat{j}) \mathrm{~m/s^2} \].

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Key Concepts

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

Uniform Circular Motion
Uniform circular motion is a mesmerizing concept where an object moves in a circular path maintaining a constant speed. This implies that while the object's speed remains unchanged, its velocity is continuously changing due to the constant change in direction.

In the context of the cat on the merry-go-round, it experiences uniform circular motion evidenced by the magnitude of its velocity being constant at both time points, even though the direction changes drastically:
  • The velocity magnitude, calculated using the Euclidean formula \( \| \vec{v} \| = \sqrt{a^2 + b^2} \) stays at \( 5.00 \ m/s \) both at \( t_1 \) and \( t_2 \).
  • The constancy of speed over time confirms that the motion remains uniform along its circular path.
Thus, in uniform circular motion, though the direction varies, the speed remains constant, characterizing the object's motion around a circle without any speeding up or slowing down.
Centripetal Acceleration
Centripetal acceleration plays a vital role in circular motion, ensuring that an object remains on its circular path. It always acts towards the center of the circle, continually changing the direction of the velocity vector.

The centripetal acceleration is determined by the formula:
  • \( a_c = \frac{v^2}{r} \)
Here, \( v \) is the speed of the object, and \( r \) is the radius of the circular path.
  • In our scenario, the calculated centripetal acceleration is approximately \( 5.24 \, \mathrm{m/s^2} \), ensuring the cat maintains its circular trajectory.
Understanding centripetal acceleration unveils how an inward force is vital for any object to execute a circular motion, effectively overcoming the object鈥檚 inertia which would otherwise propel it in a straight line.
Average Acceleration
Average acceleration measures the rate of change of velocity over a period. This is particularly interesting in circular motion as velocity changes direction, not speed. Thus, even in uniform circular motion, the object experiences average acceleration because of this directional change.

It is calculated by:
  • \( \vec{a}_{avg} = \frac{\vec{v}_2 - \vec{v}_1}{t_2-t_1} \)
Where \( \vec{v}_1 \) and \( \vec{v}_2 \) represent initial and final velocities, and \( t_2 - t_1 \) is the time interval. This calculation results in:
  • \( \vec{a}_{avg} = (-2.00 \hat{i} - 2.67 \hat{j}) \, \mathrm{m/s^2} \), depicting a directional change happening even though the speed remains constant.
By understanding average acceleration in terms of change in velocity, students can grasp how an object in uniform circular motion still experiences acceleration due to its perpetually changing direction.

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Most popular questions from this chapter

SSM ILW A particle leaves the origin with an initial velocity \(\vec{v}=(3,00 \hat{\mathrm{i}}) \mathrm{m} / \mathrm{s}\) and a constant acceleration \(\vec{a}=(-1.00 \hat{\mathrm{i}}-\) \(0 . .500 \hat{j}) \mathrm{m} / \mathrm{s}^{2}\), When it reaches its maximum \(x\) coordinate, what are its (a) velocity and (b) position vector?

A particle moves so that its position (in meters) as a function of time (in seconds) is \(\vec{r}=\hat{\mathrm{i}}+4 t \hat{\mathrm{j}}+t \hat{\mathrm{k}}\). Write cxpressions for (a) its velocity and (b) its acceleration as functions of time.

Two ships, \(A\) and \(B\), leave port at the same time. Ship \(A\) travels northwest at 24 knots, and ship \(B\) travels at 28 knots in a direction \(40^{\circ}\) west of south. ( 1 knot \(=1\) nautical mile per hour; see Appendix D.) What are the (a) magnitude and (b) direction of the velocity of ship \(A\) relative to \(B ?\) (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of \(B\) (the direction of \(B\) 's position) relative to \(A\) at that time?

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ}\), A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

A person walks up a stalled 15 -m-long escalator in \(90 \mathrm{~s}\). When standing on the same escalator, now moving, the person is carried up in \(60 \mathrm{~s}\). How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?
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