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Chapter 10: Problem 61

Starting from rest, a hollow ball rolls down a ramp inclined at angle \(\theta\) to the horizontal. Find an expression for its speed after it's gone a distance \(d\) along the incline.

Short Answer

Expert verified
The speed of the hollow ball after it has rolled down the incline for a distance \(d\) is given by \(v = \sqrt{\frac{4gdsin(\theta)}{3}}\)

Step by step solution

01

Expression for Initial Potential Energy

The initial potential energy of the ball at the top of the incline can be given by the expression \(U = mgdsin(\theta)\), where \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, \(d\) is the distance moved along the incline, and \(\theta\) is the angle of the ramp relative to horizontal.
02

Expression for Final Kinetic Energy

When the ball reaches the bottom of the incline, all of its initial potential energy is converted into kinetic energy. The total kinetic energy encompasses both translational and rotational aspects due to rolling without slipping. The kinetic energy can be expressed as \(K =\frac{1}{2}mv^2+\frac{1}{2}I\omega^2\), where \(v\) is the speed, \(I\) is the moment of inertia, and \(\omega=w=\frac{v}{R}\) is the angular velocity. For a hollow sphere, \(I =\frac{2}{3}mR^2\), where \(R\) is the radius of the ball. After substituting \(\omega\) with \(\frac{v}{R}\) and \(I\) with \(\frac{2}{3}mR^2\), we get \(K =\frac{1}{2}mv^2+\frac{1}{4}mv^2=\frac{3}{4}mv^2\)
03

Energy Conservation Equation and Solving for Final Speed

Since energy is conserved, we have \(U=K\). This conservation equation can be written as \(mgdsin(\theta)=\frac{3}{4}mv^2\). Solving for \(v\): \(v = \sqrt{\frac{4gdsin(\theta)}{3}}\) is the expression we seek for the speed of the ball after it has gone the distance \(d\) along the incline.

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