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Chapter 14: Problem 16

A small box of mass \(m\) is given a speed of \(v=\sqrt{\frac{1}{4} g r}\) at the top of the smooth half cylinder. Determine the angle \(\theta\) at which the box leaves the cylinder.

Short Answer

Expert verified
The angle of loss of contact, \(\theta\), is \(30^{\circ}\).

Step by step solution

01

Determine the Speed

Write down the given speed of the box, \(v = \sqrt{\frac{1}{4} g r}\) . This will be used later to determine the centripetal force.
02

Apply the Principle of Circular Motion

For a body in circular motion, the radial acceleration is given by \(a_r = \frac{v^2}{r}\). From this we can solve for \(v^2 = a_r r\). Now replace \(a_r\) with \(g \sin{\theta}\), representing the radial component of gravity to get \(v^2 = g r \sin{\theta}\) . Set this equation equal to the equation obtained in Step 1 to get \(\sqrt{\frac{1}{4} g r} = g r \sin{\theta}\) .
03

Determine the Angle of Loss of Contact

Solve the equation \(\sqrt{\frac{1}{4} g r} = g r \sin{\theta}\) to find the angle \(\theta\) . To solve the equation for \(\theta\), first square both sides to cancel the square root. Then divide both sides by \( g r \) to isolate \(\sin{\theta}\). Lastly, use the inverse sine function to solve for \(\theta\). This should give \(\theta = \sin^{-1}{\frac{1}{2}}\) or \(\theta = 30^{\circ}\).

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Most popular questions from this chapter

The elevator \(E\) and its freight have a total mass of \(400 \mathrm{~kg}\). Hoisting is provided by the motor \(M\) and the \(60-\mathrm{kg}\) block \(C\). If the motor has an efficiency of \(\varepsilon=0.6\), determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of \(v_{E}=4 \mathrm{~m} / \mathrm{s}\).

The \(20-\mathrm{kg}\) crate is subjected to a force having a constant direction and a magnitude \(F=100 \mathrm{~N}\). When \(s=15 \mathrm{~m}\), the crate is moving to the right with a speed of \(8 \mathrm{~m} / \mathrm{s}\). Determine its speed when \(s=25 \mathrm{~m} .\) The coefficient of kinetic friction between the crate and the ground is \(\mu_{k}=0.25\).

The escalator steps move with a constant speed of \(0.6 \mathrm{~m} / \mathrm{s}\). If the steps are \(125 \mathrm{~mm}\) high and \(250 \mathrm{~mm}\) in length, determine the power of a motor needed to lift an average mass of \(150 \mathrm{~kg}\) per step. There are 32 steps.

The block has a weight of \(80 \mathrm{lb}\) and rests on the floor for which \(\mu_{k}=0.4\). If the motor draws in the cable at a constant rate of \(6 \mathrm{ft} / \mathrm{s}\), determine the output of the motor at the instant \(\theta=30^{\circ} .\) Neglect the mass of the cable and pulleys.

To dramatize the loss of energy in an automobile, consider a car having a weight of \(5000 \mathrm{lb}\) that is traveling at \(35 \mathrm{mi} / \mathrm{h} .\) If the car is brought to a stop, determine how long a \(100-W\) light bulb must burn to expend the same amount of energy. (1 mi \(=5280 \mathrm{ft}\).)
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