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Chapter 5: Problem 54

A bowling ball weighing 71.2 \(\mathrm{N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80-\) m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the bowling ball, in magnitude and direction, at this instant? (b) What is the tension in the rope at this instant?

Short Answer

Expert verified
(a) Acceleration: 4.64 m/s虏 upwards. (b) Tension: 105.02 N.

Step by step solution

01

Understand the Pendulum Motion

The problem involves a pendulum, where the bowling ball is moving in a circular path. When it passes through the vertical, the net force provides the centripetal force necessary for the circular motion.
02

Calculate the Centripetal Acceleration

The centripetal acceleration (\( a_c \)) can be calculated using the formula: \[ a_c = \frac{v^2}{r} \]Given, \( v = 4.20 \, \text{m/s} \)and \( r = 3.80 \, \text{m} \).Substitute these values to find \( a_c \).\[ a_c = \frac{(4.20)^2}{3.80} = \frac{17.64}{3.80} \approx 4.64 \, \text{m/s}^2 \]
03

Determine the Direction of Acceleration

The acceleration at this point is centripetal, and therefore, directed towards the center of the circular path, i.e., upwards along the path of the rope.
04

Calculate the Tension in the Rope

The tension (\( T \)) in the rope must counteract the weight (\( mg \)) of the ball and provide the centripetal force. Use:\[ T = mg + ma_c \]where \( m = \frac{W}{g} = \frac{71.2}{9.8} \approx 7.27 \, \text{kg} \). Substitute \( m \), \( g = 9.8 \, \text{m/s}^2 \), and \( a_c = 4.64 \, \text{m/s}^2 \) into the equation:\[ T = 7.27 \times 9.8 + 7.27 \times 4.64 = 71.296 + 33.7248 \approx 105.02 \, \text{N} \]
05

Conclusion: Summarize Answers

(a) At the instant when the ball passes through the vertical, the acceleration of the ball is \( 4.64 \, \text{m/s}^2 \), directed upwards along the path of the rope. (b) The tension in the rope at this instant is approximately \( 105.02 \, \text{N} \).

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a fundamental concept in understanding the motion of objects moving along a circular path. It is the acceleration that points towards the center of the circle. In the context of the bowling ball problem, as it swings like a pendulum, the ball experiences centripetal acceleration at the bottom of its path.
The formula to find this type of acceleration is expressed as:
  • \( a_c = \frac{v^2}{r} \)
where \(v\) is the speed of the object and \(r\) is the radius of the circular path.
For the pendulum in question, we know:
  • Speed, \(v = 4.20 \text{ m/s} \)
  • Length of the rope (radius), \(r = 3.80 \text{ m} \)
Plugging these values into our formula, we calculate a centripetal acceleration of approximately \( 4.64 \text{ m/s}^2 \).
It's crucial to remember that this form of acceleration is always directed inwards, towards the axis of rotation, indicating that it plays a vital role in keeping the ball on its circular path.
Tension in a Rope
Tension in the rope during pendulum motion is another important aspect to understand. It refers to the force exerted along the rope as it keeps the pendulum ball attached and counterbalances the gravitational force.
Tension provides the centripetal force needed for circular motion, and its magnitude can change depending on the ball's position in the swing.
At the lowest point of the pendulum's swing, the tension is at its maximum. Here's why:
  • The rope must support the weight of the ball (\(mg\)) where \(m\) is the mass and \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).
  • An additional force due to centripetal acceleration \( (ma_c)\) is also present.
Accordingly, the expression to calculate the total tension \(T\) becomes:
  • \( T = mg + ma_c \)
Upon substituting the given values, we find that the tension in the rope when the pendulum is at the bottom of its arc is about \( 105.02 \text{ N} \). This demonstrates that tension must account for both the gravitational pull and the need for centripetal force.
Circular Motion
Circular motion describes the movement of an object along a circular path. A key aspect of this motion is that the object's velocity vector constantly changes direction, implying a continuous acceleration is needed even if the speed is constant.
In the case of the pendulum, while its path is constrained by the rope, it swings through a circular arc. Crucially, the centripetal acceleration that we calculated earlier maintains this circular path.
The swinging movement consists of:
  • Periodic motion, repeating over time.
  • Dependence on factors like the length of the rope, which influences the period of swing.
Moreover, gravity acts as the restoring force, pulling the pendulum back towards its equilibrium position, while the tension in the rope provides the necessary force to alter the direction of velocity, ensuring a curved trajectory. Understanding these principles of circular motion is essential to grasp the dynamics of the pendulum and similar systems experiencing rotational motion.

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Most popular questions from this chapter

A 5.00 -kg box sits at rest at the bottom of a ramp that is 8.00 \(\mathrm{m}\) long and that is inclined at \(30.0^{\circ}\) above the horizontal. The coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.40\) , and the coefficient of static friction is \(\mu_{\mathrm{s}}=0.50 .\) What constant force \(F,\) applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.00 \(\mathrm{s} ?\)

Force During a Jump. An average person can reach a maximum height of about 60 \(\mathrm{cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 \(\mathrm{cm}\) .To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 \(\mathrm{cm} ?\) (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(w,\) what force does the ground exert on him or her during the jump?

A box with mass \(m\) is dragged across a level floor having a coefficient of kinetic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .\) (a) In terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of the force required to move the box with constant speed. (b)Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90 -kg patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the horizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

A \(2540-\mathrm{kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by \(v(t)=\) \(A t+B t^{2},\) where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and 1.00 s later an upward velocity of 2.00 \(\mathrm{m} / \mathrm{s}\) (a) Determine \(A\) and \(B\) , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m} .\) The physics major has mass \(70.0 \mathrm{kg},\) and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere? (b) At the bottom of the circle, his speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?
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