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Chapter 8: Problem 61

A solid sphere of mass 0.600 kg rolls without slipping along a horizontal surface with a transnational speed of \(5.00 \mathrm{m} / \mathrm{s} .\) It comes to an incline that makes an angle of \(30^{\circ}\) with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?

Short Answer

Expert verified
Answer: To calculate the vertical height to which the sphere rises along the incline, follow the steps: 1. Calculate the initial total kinetic energy of the sphere. 2. Calculate the gravitational potential energy at the highest point on the incline. 3. Use the conservation of mechanical energy principle to equate the initial total kinetic energy to the gravitational potential energy at the highest point. 4. Solve for the vertical height (h) using the given values for mass, speed, and acceleration due to gravity. 5. Calculate the distance along the inclined plane using the calculated value of h and the given angle of inclination.

Step by step solution

01

Calculate the initial kinetic energy of the sphere

To do this, we need to find the total kinetic energy of the sphere, which consists of translational kinetic energy and rotational kinetic energy. For a sphere, the moment of inertia \(I = \frac{2}{5}mr^2\). The total kinetic energy, KE, is the sum of translational and rotational kinetic energy and can be found using the formula: KE_total = \(\frac{1}{2} mv^2 + \frac{1}{2} Iω^2\) Since the sphere is rolling without slipping, we can use the relation \(v = rω\). Using the given data, \(m = 0.600 \mathrm{kg}\) and \(v = 5.00 \mathrm{m} / \mathrm{s}\), we can calculate the initial total kinetic energy.
02

Calculate the gravitational potential energy at the highest point

When the sphere reaches its highest point on the incline, it comes to a stop; hence all its initial kinetic energy is transformed into gravitational potential energy. We can represent this as: PE_gravity = \(mgh\) Where \(h\) is the vertical height the sphere rises above the horizontal surface.
03

Use the conservation of mechanical energy

The conservation of mechanical energy states that the sum of kinetic and potential energies is constant. In this exercise, we can equate the initial total kinetic energy to the gravitational potential energy at the highest point to find the vertical height \(h\): KE_total = PE_gravity
04

Calculate the vertical height \(h\)

Substitute the values and equations derived from Steps 1-3, and solve for \(h\): \(\frac{1}{2} mv^2 + \frac{1}{2} Iω^2 = mgh\) \(\Rightarrow h = \frac{v^2}{2g} + \frac{Iω^2}{2mg} = \frac{v^2}{2g} + \frac{v^2}{2g} × (\frac{2}{5})\) Substitute the given values of \(m = 0.600 \mathrm{kg}\) and \(v = 5.00 \mathrm{m} / \mathrm{s}\), and use \(g = 9.81 \mathrm{m} / \mathrm{s^2}\) for the acceleration due to gravity. Solve for the vertical height \(h\).
05

Calculate the distance along the incline

Now, knowing the vertical height the sphere rises above the horizontal surface, we can find the distance along the incline by using the relation: \(h= L \sin θ\) Where \(L\) is the distance along the inclined plane and \(θ\) is the angle of inclination. Given, \(θ = 30°\), use the calculated value of \(h\) from Step 4 to calculate \(L\). Thus, following these steps, we can find the vertical height above the horizontal surface to which the sphere rises on the incline.

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Most popular questions from this chapter

Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)
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