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Chapter 2: Problem 71

A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the uverage speed and (b) the average velocity of the marble.

Short Answer

Expert verified
(a) Average speed is \(5\pi\) cm/s. (b) Average velocity is 0 cm/s.

Step by step solution

01

Identify the Problem

We need to find (a) the average speed and (b) the average velocity of a marble rolling back and forth in a hemispherical bowl. The diameter of the bowl is given, and the time for one complete journey (to the opposite rim and back) is provided.
02

Determine the Distance for Average Speed

Since the marble travels from one rim to the opposite rim twice to complete the journey, half the circumference is calculated for one way and summed up twice to give the total distance traveled by the marble. The circumference of the hemisphere is given by the formula: \\[ C = \pi \times d \] \Where \\( d = 50.0 \ \text{cm} \). Thus the distance for half a circumference is: \\[ D = \pi \times 25 \]
03

Calculate Total Distance

Since the marble goes from one rim to the opposite and back: Total distance \( D = 2 \times 25\pi \). This gives us the total distance traveled by the marble as: \\[ D = 50\pi \ \text{cm} \].
04

Compute Average Speed

Average speed is the total distance traveled divided by the total time taken. Thus, we use the formula: \\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \] \\[ \text{Average Speed} = \frac{50\pi}{10} \ = 5\pi \ \text{cm/s} \].
05

Determine Displacement for Average Velocity

The displacement of the marble is the straight-line distance between the starting point and the ending point, which, in this case, is zero since it ends up at the same starting position after 10 seconds.
06

Compute Average Velocity

The average velocity is the displacement divided by the time. Since the displacement is zero (as it comes back to the start point), the average velocity is \\[ \text{Average Velocity} = \frac{0}{10} = 0 \ \text{cm/s}. \]

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that focuses on describing the motion of objects without considering the forces causing the movement. It primarily deals with quantities such as displacement, velocity, speed, and acceleration. In the context of the marble rolling in the hemispherical bowl, kinematics allows us to analyze how the marble moves from one side of the bowl to the other and back again. This analysis includes tracking the marble鈥檚 path, considering the distance it covers, and understanding its velocity throughout the motion.
  • Displacement: Describes the change in position of the marble from its starting point to its end point, considering direction.
  • Velocity: Refers to how fast an object moves in a specific direction. Speed is the scalar quantity that clarifies how fast an object moves without directional consideration.
Circular Motion
Circular motion depicts the movement of an object along the circumference of a circle or circular path. In this exercise, the marble鈥檚 path aligns with a semicircle across the hemisphere of the bowl. Understanding circular motion helps in determining how the marble pendulates back and forth.
  • Hemispherical Path: The marble traverses a path along the bowl's rim with a known circumference.
  • Centripetal Force: While not directly calculated here, any object in circular motion, like our marble, is subject to a force pulling it toward the center of its circular path.
The marble rolls to and fro along the circular path, making use of gravity and momentum during its journey.
Displacement Calculation
Displacement calculation involves finding the shortest distance between an object's starting and final positions. It considers direction, making it a vector quantity. For the marble in the exercise, it starts and ends at the same point, meaning its displacement is zero, despite the distance it covers.
  • Vector vs Scalar: Remember, displacement is a vector quantity including direction, unlike distance, which is scalar.
  • Zero Displacement: Even when moving, if an object returns to its original point, the net displacement remains zero.
This concept is crucial in distinguishing displacement from the total path length traveled by the marble.
Distance Traveled
The distance traveled refers to the total length of the path taken by an object. Unlike displacement, distance does not consider direction, which makes it a scalar quantity. In the scenario of the marble, we calculate distance using the concept of circumference and its journey across it.
  • Circumferential Path: The journey of the marble is twice the half-circumference. Every trip from one side of the bowl to the other is a half-circumference.
  • Formula: For a full journey, distance is calculated as \( D = 50\pi \, \text{cm} \).
  • Total Path: The marble covers a significant distance as it travels to the opposite rim and back within a circular path.
Time and Motion Analysis
Time and motion analysis is the study of the time taken for an object to move from one point to another. By analyzing time along with path details, we extract meaningful parameters like speed and velocity. In this exercise, we used time to calculate both average speed and velocity of the marble, showing how time impacts motion understanding.
  • Time Factor: The duration of 10 seconds is critical in computing average speed and velocity.
  • Speed vs Velocity: Understanding their differences helps in analyzing motion. Speed is the path length per unit time, while velocity is displacement per unit time.
  • Average Calculations: Through time analysis and motion calculations, average speed and velocity reveal different insights about the motion.
Time and motion analysis enriches the understanding of how objects move through space, assisting not only in clean calculations but also in conceptual comprehension.

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Most popular questions from this chapter

Launch Failure. A \(7500-\mathrm{kg}\) rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 \(\mathrm{m} / \mathrm{s}^{2}\) and feels no approciable air resistance. When it has reached a height of \(525 \mathrm{m},\) its engines suddenly fail so that the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2} .\) Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 \(\mathrm{s}\) , Powers shuts off the engine and steps out of the helicopter Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

Touchdown on the Moon. A lunar lander is making its descent to Moon Base I (Fig. 2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 \(\mathrm{m}\) above the surface and has a downward speed of 0.8 \(\mathrm{m} / \mathrm{s}\) . With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 \(\mathrm{m} / \mathrm{s}^{2}\) .

A hot-air balloonist, rising vertically with a constant velocity of magnitude \(5.00 \mathrm{m} / \mathrm{s},\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. 2.41\()\) . After it is released, the sandbag is in frec fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds s strike after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_{y}-t\) \(v_{y^{-} t},\) and \(y-t\) graphs for the motion.

Two runners start simultaneously from the same point on a circular \(200-m\) track and \(\mathrm{run}\) in opposite directions. One runs at a constant speed of 6.20 \(\mathrm{m} / \mathrm{s}\) , and the other runs at a constant speed of 5.50 \(\mathrm{m} / \mathrm{s}\) . When they first mect, (a) for how long a time will they have been running, and (b) how far will each one have run along the track?
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