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Chapter 4: Problem 66

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=4.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

Short Answer

Expert verified
The center of the circular path is at coordinates (2.00 m, 6.00 m).

Step by step solution

01

Understand the Motion

The motion describes a particle moving along a circular path at a constant speed. This indicates that the particle experiences centripetal acceleration directed towards the center of the circle, and its velocity changes direction but not magnitude.
02

Analyze Velocities and Accelerations

At time \( t_1 = 4.00 \) s, the velocity is \((3.00 \, \text{m/s}) \hat{\mathbf{j}}\) and acceleration is in the positive \(x\)-direction. At \( t_2 = 10.0 \) s, the velocity is \((-3.00 \, \text{m/s}) \hat{\mathbf{i}}\) and acceleration is in the positive \(y\)-direction. This suggests a quarter-circle of motion between these times.
03

Define Velocity and Position Relation

For a circular motion path with constant speed, the direction of velocity and the position with respect to the center of the circle is perpendicular. When a particle moves from \((3.00 \, \text{m/s}) \hat{\mathbf{j}}\) to \((-3.00 \, \text{m/s}) \hat{\mathbf{i}}\), it indicates that each velocity vector is radial to the center.
04

Determine Center from Velocity Vectors

Since velocities are perpendicular corresponding to circular motion's radii quadrants, we can infer the velocity change involves a rotation from \( \hat{\mathbf{j}} \) to \(-\hat{\mathbf{i}} \). The center of the circle must be equidistant from both the velocity's endpoint and the original center. This implies a circle centered to accommodate these vectors' directions.
05

Calculate Displacement to the Center

Given the point \((5.00 \, \text{m}, 6.00 \, \text{m})\) at \( t_1 \), we find that displacement from this position due to velocity \( (3.00 \, \text{m/s}) \hat{\mathbf{j}} \) to the center must be \(-3.00 m \) unit left \((- \hat{\mathbf{i}})\). Therefore, the center is \((5.00 \, \text{m} - 3.00 \, \text{m}, 6.00 \, \text{m}) = (2.00 \, \text{m}, 6.00 \, \text{m})\).
06

Calculate Center with Other Quadrant Displacement

From the point at \((t2)\) the displacement, given velocity is \((-3.00 \, \text{m/s}) \hat{\mathbf{i}} \) negative x, implies adjustment \(0\) units up, thus staying at \((2.00 \, \text{m}, 6.00 \, \text{m})\). The confirmed center is consistent with circular motion constraints.

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Key Concepts

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

Constant Speed
In circular motion, when a particle moves at constant speed, it isn't the speed that changes but the direction of the motion. This means the particle's velocity changes due to its direction, while the magnitude remains constant. Constant speed signifies the uniform motion of the particle around the circular path.
  • The magnitude of the velocity remains the same.
  • The direction of the velocity is always tangential to the circle.
Due to the consistent speed along the circular path, the particle requires an inward force to change direction, leading to centripetal acceleration.
Centripetal Acceleration
Centripetal acceleration is the key factor that keeps a particle moving in a circular path. It always points towards the center of the circle, causing the change in direction of the velocity vector despite a constant speed.
  • It is calculated using the formula: \[ a_c = \frac{v^2}{r} \]where \( v \) is the constant speed and \( r \) is the radius of the circle.
  • This acceleration is not to be confused with tangential acceleration, which affects the speed rather than direction.
  • In the given exercise, the acceleration at different times also helps to confirm the orientation and position related to the circle's center.
Centripetal acceleration is crucial for understanding how the change in velocity direction happens without changing speed.
Velocity Vectors
In circular motion, velocity vectors provide insight into the movement at any moment. These vectors are always tangent to the path of motion.Notice how, in the problem, the velocity changes:
  • At time \( t_1 \), the velocity vector is in the positive \(y\)-direction, \( (3.00 \, \text{m/s}) \hat{\mathbf{j}} \).
  • At time \( t_2 \), it switches to the negative \(x\)-direction, \( (-3.00 \, \text{m/s}) \hat{\mathbf{i}} \).
These changes highlight the quarter-circle movement, illustrating how velocity vectors rotate as the particle progresses around the circle. The radial orientation of these vectors to the circle's center helps deduce the circle's size and position.
Motion Analysis
Analyzing the motion of a particle in a circular path involves understanding both the positional changes and the velocity alterations.The exercise demonstrates how to determine the circle's center using the velocity orientations:
  • The transition from \( (3.00 \, \text{m/s}) \hat{\mathbf{j}} \) to \( (-3.00 \, \text{m/s}) \hat{\mathbf{i}} \) over a time gap suggests a quarter-circle.
  • By finding the common center for these vectors, we establish the circle's coordinates.
  • The calculated center \( (2.00 \, \text{m}, 6.00 \, \text{m}) \) ensures both vectors are consistently radial.
Such analysis is valuable in understanding and visualizing circular motion in physics problems, confirming the effects of constant speed and directional changes.

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Most popular questions from this chapter

A baseball is hit at Fenway Park in Boston at a point \(0.762 \mathrm{~m}\) above home plate with an initial velocity of \(33.53 \mathrm{~m} / \mathrm{s}\) directed \(55.0^{\circ}\) above the horizontal. The ball is observed to clear the \(11.28\) -m-high wall in left field (known as the "green monster") \(5.00 \mathrm{~s}\) after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate \(0.500 \mathrm{~s}\) before it clears the wall.

A frightened rabbit moving at \(6.00 \mathrm{~m} / \mathrm{s}\) due east runs onto a large area of level ice of negligible friction. As the rabbit slides across the ice, the force of the wind causes it to have a constant acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2}\), due north. Choose a coordinate system with the origin at the rabbit's initial position on the ice and the positive \(x\) axis directed toward the east. In unit-vector notation, what are the rabbit's (a) velocity and (b) position when it has slid for \(3.00 \mathrm{~s}\) ?

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{i}+4 t \hat{j}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis?

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. \((1 \mathrm{knot}=1\) 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 moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?
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