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Chapter 8: Problem 59

A solid sphere is released from rest and allowed to roll down a board that has one end resting on the floor and is tilted at \(30^{\circ}\) with respect to the horizontal. If the sphere is released from a height of \(60 \mathrm{cm}\) above the floor, what is the sphere's speed when it reaches the lowest end of the board?

Short Answer

Expert verified
Answer: The sphere's speed when it reaches the lowest end of the board is approximately \(2.45 \thinspace \mathrm{m/s}\).

Step by step solution

01

Finding the distance traveled along the board

We can find the distance traveled, denoted as 'd', with the help of trigonometry. Since we know the angle \(\theta = 30^{\circ}\) and the height \(h = 60 \thinspace \mathrm{cm}\), we can use the sine function to find 'd': \(\sin\theta = \frac{h}{d}\) Now we need to find 'd': \(d = \frac{h}{\sin\theta}\)
02

Converting height to meters

To keep the units consistent and SI, we need to convert the height to meters: \(h = 60 \thinspace \mathrm{cm} * \frac{1 \thinspace \mathrm{meter}}{100 \thinspace \mathrm{cm}} = 0.6 \thinspace \mathrm{m}\)
03

Calculate the distance 'd'

Now we can plug in the values of \(h\) and \(\theta\) to find 'd': \(d = \frac{0.6 \thinspace \mathrm{m}}{\sin{30^{\circ}}}=1.2 \thinspace \mathrm{m}\)
04

Apply conservation of energy

The sum of the kinetic and potential energies of the sphere remains constant throughout its motion. At the top, the sphere has potential energy, and at the bottom, it has kinetic energy (the rolling kinetic energy includes the linear and rotational kinetic energy). At the top: \(E_\text{potential} = mgh\) At the bottom: \(E_\text{linear} = \frac{1}{2}mv^2\) \(E_\text{rotational} = \frac{2}{5}mv^2\) (Since a solid sphere has moment of inertia \(\frac{2}{5}mr^2\)) Now we equate the potential energy at the top with the sum of the linear and rotational kinetic energies at the bottom \(mgh = \frac{1}{2}mv^2 + \frac{2}{5}mv^2\)
05

Solve for the sphere's speed v

We can simplify the equation by dividing both sides by \(m\) and combining the terms for \(v^2\): \(gh = (\frac{1}{2} + \frac{2}{5})v^2\) We can now solve for \(v^2\): \(v^2 = \frac{gh}{(\frac{1}{2} + \frac{2}{5})}\) And finally, we take the square root to find 'v': \(v = \sqrt{\frac{gh}{(\frac{1}{2} + \frac{2}{5})}}\)
06

Final calculation

Plug in the values for \(g = 9.81 \thinspace \mathrm{m/s^2}\) and \(h = 0.6 \thinspace \mathrm{m}\) to calculate the speed of the sphere when it reaches the bottom: \(v = \sqrt{\frac{(9.81 \thinspace \mathrm{m/s^2})(0.6 \thinspace \mathrm{m})}{(\frac{1}{2} + \frac{2}{5})}} \approx 2.45 \thinspace \mathrm{m/s}\) Therefore, the sphere's speed when it reaches the lowest end of the board is approximately \(2.45 \thinspace \mathrm{m/s}\).

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