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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 student of mass \(52 \mathrm{~kg}\) wants to measure the mass of a playground merry-go-round, which consists of a solid metal disk of radius \(R=1.5 \mathrm{~m}\) that is mounted in a horizontal position on a low-friction axle. She tries an experiment: She runs with speed \(v=6.8 \mathrm{~m} / \mathrm{s}\) toward the outer rim of the merry-go-round and jumps on to the outer rim, as shown in the figure. The merry-go-round is initially at rest before the student jumps on and rotates at \(1.3 \mathrm{rad} / \mathrm{s}\) immediately after she jumps on. You may assume that the student's mass is concentrated at a point. a) What is the mass of the merry-go-round? b) If it takes 35 s for the merry-go-round to come to a stop after the student has jumped on, what is the average torque due to friction in the axle? c) How many times does the merry-go-round rotate before it stops, assuming that the torque due to friction is constant?

A disk of clay is rotating with angular velocity \(\omega .\) A blob of clay is stuck to the outer rim of the disk, and it has a mass \(\frac{1}{10}\) of that of the disk. If the blob detaches and flies off tangent to the outer rim of the disk, what is the angular velocity of the disk after the blob separates? a) \(\frac{5}{6} \omega\) b) \(\frac{10}{11} \omega\) c) \(\omega\) d) \(\frac{11}{10} \omega\) e) \(\frac{6}{5} \omega\)

-10.62 The Earth has an angular speed of \(7.272 \cdot 10^{-5} \mathrm{rad} / \mathrm{s}\) in its rotation. Find the new angular speed if an asteroid \(\left(m=1.00 \cdot 10^{22} \mathrm{~kg}\right)\) hits the Earth while traveling at a speed of \(1.40 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\) (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases: a) The asteroid hits the Earth dead center. b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation. c) The asteroid hits the Earth nearly tangentially in the direction opposite to Earth's rotation.

Using the conservation of mechanical energy, calculate the final speed and the acceleration of a cylindrical object of mass \(M\) and radius \(R\) after it rolls a distance \(s\) without slipping along an inclined plane of angle \(\theta\) with respect to the horizontal

You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 20.0 -m-tall building and land on the ground safely at a final vertical speed of \(4.00 \mathrm{~m} / \mathrm{s}\). At the edge of the building's roof, there is a \(100 .-\mathrm{kg}\) drum that is wound with a sufficiently long rope (of negligible mass), has a radius of \(0.500 \mathrm{~m}\), and is free to rotate about its cylindrical axis with a moment of inertia \(I_{0}\). The script calls for the 50.0 -kg stuntman to tie the rope around his waist and walk off the roof. a) Determine an expression for the stuntman's linear acceleration in terms of his mass \(m\), the drum's radius \(r\) and moment of inertia \(I_{0}\). b) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of \(4.00 \mathrm{~m} / \mathrm{s},\) and use this value to calculate the moment of inertia of the drum about its axis. c) What is the angular acceleration of the drum? d) How many revolutions does the drum make during the fall?
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