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Chapter 10: Problem 9

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first? a) The solid sphere arrives first. b) The box arrives first. c) Both arrive at the same time. d) It is impossible to determine.

Short Answer

Expert verified
(a) The solid sphere or (b) The box? Answer: (b) The box arrives first.

Step by step solution

01

Analyze the motion of the box

The box slides down the incline with no friction, so it means its motion follows standard free-fall kinematics, without rotational kinetic energy. The acceleration of the box, which is given by Newton's second law, can be expressed as: \( a_{box} = g \cdot \sin(\theta) \) where \(a_{box}\) is the acceleration of the box, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of the incline.
02

Analyze the motion of the solid sphere

As the sphere is rolling down the incline without slipping, we need to consider both its linear and angular components of motion. For the sphere, total kinetic energy is a combination of translational (linear) and rotational (angular) kinetic energy. Therefore, the acceleration of the solid sphere can be expressed as: \( a_{sphere} = \frac{5}{7} g \cdot \sin(\theta) \) This equation represents the acceleration of the solid sphere, considering the no-slipping condition and the specific moment of inertia for a solid sphere which is \(\frac{2}{5} mR^2\).
03

Compare the accelerations of the box and solid sphere

Now that we have the expressions for the accelerations of both the box and the solid sphere, we can compare them: \( a_{box} = g \cdot \sin(\theta) \) \( a_{sphere} = \frac{5}{7} g \cdot \sin(\theta) \) We see that the acceleration of the box is greater than the acceleration of the solid sphere, so the box will reach the bottom of incline faster than the solid sphere.
04

Answer

The correct answer is (b) The box arrives first.

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Most popular questions from this chapter

A figure skater draws her arms in during a final spin. Since angular momentum is conserved, her angular velocity will increase. Is her rotational kinetic energy conserved during this process? If not, where does the extra energy come from or go to?

An object made of two disk-shaped sections, \(\mathrm{A}\) and \(\mathrm{B}\), as shown in the figure, is rotating about an axis through the center of disk A. The masses and the radii of disks \(A\) and \(B\). respectively are, \(2.00 \mathrm{~kg}\) and \(0.200 \mathrm{~kg}\) and \(25.0 \mathrm{~cm}\) and \(2.50 \mathrm{~cm}\). a) Calculate the moment of inertia of the object. b) If the axial torque due to friction is \(0.200 \mathrm{~N} \mathrm{~m}\), how long will it take for the object to come to a stop if it is rotating with an initial angular velocity of \(-2 \pi \mathrm{rad} / \mathrm{s} ?\)

In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

A sheet of plywood \(1.3 \mathrm{~cm}\) thick is used to make a cabinet door \(55 \mathrm{~cm}\) wide by \(79 \mathrm{~cm}\) tall, with hinges mounted on the vertical edge. A small 150 - \(\mathrm{g}\) handle is mounted \(45 \mathrm{~cm}\) from the lower hinge at the same height as that hinge. If the density of the plywood is \(550 \mathrm{~kg} / \mathrm{m}^{3},\) what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.
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