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Chapter 6: Problem 55

As your bus rounds a flat curve at constant speed, a package with mass \(0.500 \mathrm{kg},\) suspended from the luggage compartment of the bus by a string 45.0 \(\mathrm{cm}\) long, is found to hang at rest relative to the bus, with the string making an angle of \(30.0^{\circ}\) with the vertical. In this position, the package is 50.0 \(\mathrm{m}\) from the center of curvature of the curve. What is the speed of the bus?

Short Answer

Expert verified
The speed of the bus is approximately 12.8 m/s.

Step by step solution

01

Understand the Scenario

We have a bus taking a curve and a package hanging inside the bus. The string suspending the package makes an angle of 30 degrees with the vertical, indicating a centripetal force acting on the package due to the curve.
02

Identify the Forces

The forces acting on the package are tension in the string and gravity. The tension has two components: one vertical (balancing gravity) and one horizontal (providing the centripetal force).
03

Resolve the Tension into Components

The tension (T) can be resolved into:- Vertical component: \(T_y = T \cos(30^{\circ})\)- Horizontal component providing centripetal force: \(T_x = T \sin(30^{\circ})\)
04

Apply Newton's Second Law Vertically

Since the package is at rest vertically: \\[ T \cos(30^{\circ}) = mg \] \where \(m = 0.500\, \text{kg}\).
05

Apply Newton's Second Law Horizontally

The horizontal component of tension provides the centripetal force needed for circular motion: \\[ T \sin(30^{\circ}) = \frac{m v^2}{r} \] \where \(r = 50.0\, \text{m}\).
06

Combine Equations

From the vertical force balance, \\[ T = \frac{mg}{\cos(30^{\circ})} \] \Substitute into the horizontal force equation to find \(v\): \\[ \frac{mg \sin(30^{\circ})}{\cos(30^{\circ})} = \frac{m v^2}{r} \]
07

Solve for the Speed

Cancel \(m\) and solve for \(v\): \[ v^2 = rg \tan(30^{\circ}) \] \[ v = \sqrt{50.0 \times 9.81 \times \frac{1}{\sqrt{3}}} \]Calculating, \v \approx 12.8 \, \text{m/s}.

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Key Concepts

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

Centripetal Force
Centripetal force is the force that acts on an object moving in a circular path and is directed towards the center around which the object is moving. It is essential for keeping the object in motion along the circular path. When a package is hanging inside a bus moving around a curve, it experiences this centripetal force.
In the given scenario, the centripetal force is not an additional force but rather comes from the horizontal component of the tension in the string. Essentially:
  • The object (package) tends to move in a straight line due to inertia.
  • The tension in the string provides the needed inward or centripetal force to keep it moving in a circular path.
  • This force ensures that the package stays in its circular trajectory parallel to the curve taken by the bus.
Understanding centripetal force helps reveal why the package hangs at a specific angle in relation to the vertical while the bus navigates the curve.
Newton's Second Law
Newton's Second Law states that the force acting on an object is equal to the mass of that object times its acceleration (\[ F = ma \]). It's crucial for analyzing the forces at play in any physics problem involving motion. In the context of circular motion:
  • The acceleration is directed towards the center of the circular path (centripetal acceleration).
  • The centripetal force can be expressed as \[ F_c = rac{mv^2}{r} \], where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the curve.
  • This equation clarifies how changes in speed or radius affect the required centripetal force.
Newton's Second Law is key to establishing the relationship between the forces we observe鈥攖ension, gravity鈥攁nd the centripetal force required for circular motion. By resolving the forces in both vertical and horizontal components, it helps further understand how the package behaves in the scenario.
Trigonometry in Physics
Trigonometry is often used in physics to resolve forces into their components. In this exercise, the angle between the string and the vertical direction calls for trigonometric functions:
  • Cosine and sine functions help break down the tension into vertical and horizontal components.
  • The vertical component \( T_y = T \cos(30^{\circ}) \) balances the weight of the package, \( mg \).
  • The horizontal component \( T_x = T \sin(30^{\circ}) \) supplies the centripetal force needed for the circular motion.
Trigonometry allows us to solve for unknowns indirectly by relating angles and side lengths. The use of \[\tan(30^{\circ})\] was crucial for finding the bus's speed by linking it to the radius and gravitational force. This mathematical approach simplifies complex problems by breaking down forces into simpler, calculable parts.

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Most popular questions from this chapter

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the plattorm when it is brought up to a speed of 40.0 \(\mathrm{rev} / \mathrm{min}\) , provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

Effect on blood of walking. While a person is walking, his arms swing through approximately a \(45^{\circ}\) angle in \(\frac{1}{2}\) s. As a reasonable approximation, we can assume that the arm moves with constant speed during each swing. A typical arm is 70.0 \(\mathrm{cm}\) long, measured from the shoulder joint. (a) What is the acceleration of a 1.0 gram drop of blood in the fingertips at the bottom of the swing? (b) Make a free-body diagram of the drop of blood in part (a).(c) Find the force that the blood vessel must exert on the drop of blood in part (b). Which way does this force point? (d) What force would the blood vessel exert if the arm were not swinging?

Stunt pilots and fighter pilots who fly at high speeds in a downward-curving arc may experience a "red out," in which blood is forced upward into the flier's head, potentially swelling or breaking capillaries in the eyes and leading to a reddening of vision and even loss of consciousness. This effect can occur at centripetal accelerations of about 2.5\(g^{\prime}\) s. For a stunt plane flying at a speed of 320 \(\mathrm{km} / \mathrm{h}\) , what is the minimum radius of downward curve a pilot can achieve without experiencing a red out at the top of the arc? (Hint: Remember that gravity provides part of the centripetal acceleration at the top of the arc; it's the acceleration required in excess of gravity that causes this problem.)

An 8.00 -kg point mass and a 15.0 -kg point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the 8.00 -kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

The Indy 500 . The Indianapolis Speedway (home of the Indy 500 ) consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at \(9^{\circ} 12^{\prime} .\) What is the no-friction-needed speed (in \(\mathrm{m} / \mathrm{s}\) and mph) for these turns? (Do you think drivers actually take the turns at that speed?
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