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A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

Short Answer

Expert verified
The initial speed of the ball was approximately 2.68 m/s.

Step by step solution

01

Understand the Problem

We need to find the initial speed of the ball as it starts moving up the ramp. The ball possesses both translational and rotational kinetic energy at the beginning, which gets converted entirely into gravitational potential energy when it stops momentarily at its maximum height on the ramp.
02

Write Down Energy Conservation Equation

Since energy is conserved, the initial kinetic energy should equal the potential energy at the maximum height. For a solid ball, the initial kinetic energy is the sum of its translational and rotational kinetic energy: \[ KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2, \] where \( I = \frac{2}{5}mr^2 \) for a solid sphere and \( \omega = \frac{v}{r} \).
03

Rearrange the Energy Equation

Substitute the expression for \( I \) and \( \omega \) into the kinetic energy equation: \[ KE = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2. \] This is the total initial kinetic energy.
04

Set Gravitational Potential Energy Equation

Determine the gravitational potential energy (PE) at the top of the ramp: \[ PE = mgh, \] where \( h \) (the height) is \( d \sin(\theta) \) with \( d = 1.50 \) m and \( \theta = 15.0^{\circ} \).
05

Equate Total Initial Kinetic Energy and Gravitational Potential Energy

Set the initial kinetic energy equal to the gravitational potential energy: \[ \frac{7}{10}mv^2 = mgd\sin(\theta). \] Simplify and solve for \( v \): \[ v^2 = \frac{10}{7} \cdot g \cdot 1.50 \cdot \sin(15^{\circ}), \] \[ v = \sqrt{\frac{10}{7} \cdot 9.8 \cdot 1.50 \cdot \sin(15^{\circ})}. \]
06

Calculate the Initial Speed

Substitute the known values into the expression and compute \( v \): \[ v = \sqrt{\frac{10}{7} \cdot 9.8 \cdot 1.50 \cdot \sin(15^{\circ})} \approx 2.68 \text{ m/s}. \]

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is the energy due to an object's rotation. Think of it as the energy a spinning object possesses because of its motion. When dealing with rotating bodies, like a rolling ball, this kind of energy is crucial to understand. The rotational kinetic energy is given by the formula: - \[ KE_{rot} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia (a measure of an object's resistance to changes in its rotation) and \( \omega \) is the angular velocity. For a solid sphere like a ball, the moment of inertia \( I \) is \( \frac{2}{5} mr^2 \), while \( \omega = \frac{v}{r} \), where \( v \) is the linear speed and \( r \) is the radius of the sphere. In our exercise, the rotational kinetic energy is part of the ball's total initial energy. When solving problems involving rolling objects, it's important to include both translational (straight-line motion) and rotational kinetic energy. As the ball starts to climb the ramp, this energy gets gradually converted into gravitational potential energy, a process central to understanding energy conservation in physics.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object has due to its position in a gravitational field. When you lift an object, you're giving it the potential to do work due to gravity pulling it back down. This potential energy is described by the equation: - \[ PE = mgh \] where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) near Earth's surface), and \( h \) is the height above the reference point. In our exercise, the ball rolls up an incline, converting its kinetic energy into gravitational potential energy as it gains height. At the peak point when the ball momentarily stops, all its initial kinetic energy has transformed into gravitational potential energy. This energy conversion helps us understand the maximum altitude reached by the ball, which allows us to determine its initial speed.
Kinematic Equations
Kinematic equations are the mathematical formulas describing the motion of objects. They relate various parameters of motion like velocity, acceleration, time, and displacement. While the detailed kinematic formulas weren't explicitly necessary for this rolling ball problem, they help outline complex motion situations where accelerations and time factors are involved. In this exercise, the focus wasn't solely on kinematic equations but on using the concept of displacement through the ramp to calculate the gravitational potential energy: - 1. The height \( h \) of the ramp is obtained using the horizontal distance \( d \) and the incline angle \( \theta \): \[ h = d\sin(\theta) \] 2. This relation aids in applying gravitational potential energy equations effectively. While kinematic equations might not have been the centerpiece of finding the initial speed in this particular problem, understanding displacement and trigonometric relations helps in navigating such scenarios.

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Most popular questions from this chapter

At one instant, force \(\vec{F}=4.0 \mathrm{j} \mathrm{N}\) acts on a \(0.25 \mathrm{~kg}\) object that has position vector \(\vec{r}=(2.0 \mathrm{i}-2.0 \mathrm{k}) \mathrm{m}\) and velocity vector \(\vec{v}=(-5.0 \hat{i}+5.0 \mathrm{k}) \mathrm{m} / \mathrm{s}\). About the origin and in unit-vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?

The rotor of an electric motor has rotational inertia \(I_{m}=\) \(2.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia \(I_{p}=12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about this axis. Calculate the number of revolutions of the rotor required to turn the probe through \(30^{\circ}\) about its central axis.

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. \(11-38\) ) slides on the lane with initial speed \(v_{\mathrm{com}}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0 .\) The coefficient of kinetic friction between the ball and the lane is \(0.21 .\) The kinetic frictional force \(\vec{f}_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\operatorname{com}}\) in terms of \(\omega ?\) During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

A ballerina begins a tour jeté (Fig. \(11-19 a\) ) with angular speed \(\omega_{i}\) and a rotational inertia consisting of two parts:\(I_{\mathrm{leg}}=1.44 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for her leg extended outward at angle \(\theta=90.0^{\circ}\) to her body and \(I_{\text {trunk }}=0.660 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for the rest of her body (primarily her trunk ). Near her maximum height she holds both legs at angle \(\theta=30.0^{\circ}\) to her body and has angular speed \(\omega_{f}(\) Fig. \(11-19 b)\). Assuming that \(I_{\text {trunk }}\) has not changed, what is the ratio \(\omega_{f} / \omega_{i} ?\)

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