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A ball swings in a vertical circle at the end of a rope \(1.50 \mathrm{m}\) long. When the ball is \(36.9^{\circ}\) past the lowest point on its way up, its total acceleration is \((-22.5 \hat{\mathbf{i}}+20.2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) At that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.

Short Answer

Expert verified
The radial acceleration of the ball is \(30.2 m/s^2\), the speed is \(6.71 m/s\), and the velocity is \(6.71 m/s\) at an angle of \(36.9^{\circ}\) upwards to the right.

Step by step solution

01

Understanding the Components of Acceleration

The total acceleration of the ball is given as \(-22.5 \hat{\mathbf{i}}+20.2 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}^{2}\). The components are in \(\hat{\mathbf{i}}\) (horizontal) and \(\hat{\mathbf{j}}\) (vertical), respectively. The acceleration due to gravity acts in the negative \(\hat{\mathbf{j}}\) direction, so its tangential acceleration is the acceleration in the \(\hat{\mathbf{i}}\) direction, or \(-22.5 \mathrm{m/s^2}\), and the acceleration due to gravity is \(20.2 \mathrm{m/s^2}\)
02

Calculating Radial Acceleration

The radial (centripetal) acceleration can be calculated using the equation \(a_r = \sqrt{{a_t}^2 + g^2} = \sqrt{(-22.5 m/s^2)^2 + (20.2 m/s^2)^2} = 30.2 m/s^2\). The negative sign of the tangential acceleration indicates that the speed of the ball is decreasing.
03

Calculating Speed

Let's determine the speed of the ball. The sum of squares of radial and tangential acceleration gives the total acceleration. We know that radial acceleration \(a_r = v^2/ r\). Therefore, we can solve for \(v = \sqrt{a_r * r}\). By substituting the values, we get \(v = \sqrt{30.2 m/s^2 * 1.5 m} = 6.71 m/s\).
04

Calculating Velocity and Direction

The velocity of the ball can be found using the formula \(v = r \omega\), where \( \omega\) is the angular velocity. Given that the ball's radial acceleration is \(30.2 m/s^2\), we can solve for \( \omega\) by using the formula \( a_r = r \omega^2\). Solving for \( \omega\) gives \( \omega = \sqrt{a_r / r} = \sqrt{30.2 m/s^2 / 1.5 m} = 4.47 rad/s\). Now, the velocity of the ball can be calculated as \(v = 1.5m * 4.47 rad/s = 6.71 m/s\). The direction of the velocity is tangential to the circle, which in this case will be upward and to the right, at an angle of \(36.9^{\circ}\), corresponding to the position of the ball.

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Key Concepts

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

Radial Acceleration
In circular motion, radial acceleration, also known as centripetal acceleration, is what keeps an object moving along a curved path. It is directed towards the center of the circle. This means it's always aiming inward, pulling the object toward the center to maintain its circular path.

For example, when a ball is swinging in a vertical circle, the radial acceleration is crucial. Its formula is given by: \[ a_r = \frac{v^2}{r} \] Here, \( a_r \) is the radial acceleration, \( v \) is the linear speed, and \( r \) is the radius of the circle.

This formula tells us that radial acceleration depends on how fast the object is going and how far the path curves (i.e., the radius). In our problem:
  • We calculated \( a_r \) as approximately \(30.2 \ \mathrm{m/s^2}\).
  • This value helps us find both the speed and direction of the ball along its path.

It's fascinating how this inward pull allows objects to curve instead of flying away in a straight line. Understanding this will help in recognizing how forces come into play in the world of circular motion.
Tangential Acceleration
Tangential acceleration is the component of acceleration that is parallel to the direction of velocity. This affects the speed of the object as it moves along its circular path. It either speeds up or slows down the object, depending on the situation.

In our scenario of a ball swinging on a rope, tangential acceleration is key for understanding changes in speed:
  • The horizontal component, \(-22.5 \ \mathrm{m/s^2}\), represents the tangential acceleration in this problem.
  • A negative sign indicates that the ball is slowing down as it moves upward.
Keep in mind that tangential acceleration is caused by external forces like gravity. These forces' effects can change depending on the object's position along its circular path.
To find the overall acceleration experienced by the object, tangential and radial accelerations are combined using the Pythagorean theorem. This blend of forces is what you feel when swinging on a rope and experiencing changes in motion.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a point or axis. It's an important concept when discussing objects in circular motion, such as our swinging ball.

Angular velocity is calculated using the relationship between radial acceleration and radius:
  • The formula to find angular velocity \( \omega \) is \[ \omega = \sqrt{\frac{a_r}{r}} \]
  • In this problem, we determined \( \omega \) to be approximately \( 4.47 \ \mathrm{rad/s} \).
This measure helps in understanding the object's rotational speed. Unlike linear speed, angular velocity tells us how fast the angle is changing as the object swings.
The computed radial and tangential accelerations help pinpoint the exact speed and angle direction in the rotation, offering a comprehensive view of the motion. Embracing this concept aids in predicting movement patterns in various real-world scenarios, enriched by a deeper grasp of physics principles in circular motion.

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Most popular questions from this chapter

A quarterback throws a football straight toward a receiver with an initial speed of \(20.0 \mathrm{m} / \mathrm{s},\) at an angle of \(30.0^{\circ}\) above the horizontal. At that instant, the receiver is \(20.0 \mathrm{m}\) from the quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

A motorist drives south at \(20.0 \mathrm{m} / \mathrm{s}\) for \(3.00 \mathrm{min}\) then turns west and travels at \(25.0 \mathrm{m} / \mathrm{s}\) for \(2.00 \mathrm{min},\) and finally travels northwest at \(30.0 \mathrm{m} / \mathrm{s}\) for 1.00 min. For this 6.00 -min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive \(x\) axis point east.

Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, \(21.0 \mathrm{m}\) high, located \(130 \mathrm{m}\) from home plate. The ball is hit at an angle of \(35.0^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time at which the ball reaches the cheap seats, and (c) the velocity components and the speed of the ball when it passes over the top row. Assume the ball is hit at a height of \(1.00 \mathrm{m}\) above the ground.

Two swimmers, Alan and Beth, start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed \(c(c>v),\) relative to the water. Alan swims downstream a distance \(L\) and then \(\mathrm{up}^{-}\) stream the same distance. Beth swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance \(L\) and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? (Note: First guess the answer.)

The water in a river flows uniformly at a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\) between parallel banks \(80.0 \mathrm{m}\) apart. You are to deliver a package directly across the river, but you can swim only at \(1.50 \mathrm{m} / \mathrm{s} .\) (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) What If? If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried?

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