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(II) A sphere of radius \(r = 34.5 cm\) and mass \(m = 1.80 kg\) starts from rest and rolls without slipping down a 30.0\(^{\circ}\) incline that is 10.0 m long. \((a)\) Calculate its translational and rotational speeds when it reaches the bottom. \((b)\) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: \((c)\) do your answers in \((a)\) and \((b)\) depend on the radius of the sphere or its mass?

Short Answer

Expert verified
(a) Translational speed: 8.64 m/s, Rotational speed: 25.1 rad/s. (b) KE ratio: 2.52. (c) Speeds and ratios do not depend on radius or mass.

Step by step solution

01

Analyze the Motion

The sphere rolls without slipping down the incline, which means that its translational motion is directly related to its rotational motion. We know the incline length is 10.0 meters and angle is 30 degrees.
02

Apply Energy Conservation

Initially, all energy is potential: \[ E_p = mgh = mgL\sin(\theta) \]At the bottom, this potential energy is converted into translational kinetic energy \( K_{trans} \) and rotational kinetic energy \( K_{rot} \): \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For a sphere, the moment of inertia \( I = \frac{2}{5}mr^2 \), and since it rolls without slipping \( v = r\omega \).
03

Solve for Translational Speed

Substitute \( \omega = \frac{v}{r} \) and \( I = \frac{2}{5}mr^2 \) into the energy conservation equation: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 \] Simplify equation: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \] Solve for \( v \): \[ v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10gL\sin(\theta)}{7}} \] Plug in numbers: \[ v = \sqrt{\frac{10 \times 9.8 \times 10 \times \sin(30^{\circ})}{7}} \] \[ v = 8.64 \text{ m/s (Translational Speed)} \]
04

Solve for Rotational Speed

Using \( v = r\omega \), solve for \( \omega \): \[ \omega = \frac{v}{r} = \frac{8.64}{0.345} \approx 25.1 \text{ rad/s (Rotational Speed)} \]
05

Calculate Kinetic Energy Ratio

Use the kinetic energy relations, \( K_{trans} = \frac{1}{2}mv^2 \) and \( K_{rot} = \frac{1}{2}I\omega^2 \):For translational:\[ K_{trans} = \frac{1}{2} \times 1.8 \times (8.64)^2 \approx 67.4 \text{ J} \]For rotational:\[ K_{rot} = \frac{1}{2} \times \frac{2}{5} \times 1.8 \times (0.345)^2 \times 25.1^2 \approx 26.8 \text{ J} \]Calculate the ratio:\[ \frac{K_{trans}}{K_{rot}} = \frac{67.4}{26.8} \approx 2.52 \]
06

Consider Dependence on Radius and Mass

Observe that the final translational speed \( v = \sqrt{\frac{10gL\sin(\theta)}{7}} \) and rotational speed \( \omega = \frac{v}{r} \) do not depend on mass \( m \). The speed is determined by the gravitational pull, length of the incline, and angle. The radius cancels in the ratio calculation, making the responses independent of the sphere's radius.

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Key Concepts

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

Rotational Motion
In understanding rotational motion, it's important to recognize how an object spins around a point or axis. When it comes to our rolling sphere, its motion consists not just of sliding down the incline, but also of spinning around its own center.

The unique aspect of rotational motion is characterized by its angular velocity, denoted by \( \omega \), which indicates how quickly the object is rotating. For rolling spheres like ours, there's a relationship between linear velocity \( v \) and angular velocity \( \omega \) described by the formula \( v = r\omega \), where \( r \) is the sphere's radius. This means that the faster the sphere rolls, the quicker it spins.

To find the rotational speed at the bottom of the incline, we use this relationship, solving for \( \omega \) as \( \omega = \frac{v}{r} \). Given that we know \( v \) from earlier steps, calculating \( \omega \) is straightforward. This shows that rotational motion is an intrinsic component of the sphere’s overall movement.
Translational Motion
Translational motion happens when an object moves from one location in space to another. Consider the sphere as it rolls down the incline, it travels a certain distance due to gravitational force pulling it along.

This type of motion is characterized by its linear velocity \( v \). We can compute \( v \) using energy principles, specifically the conservation of energy which equates the initial potential energy of the sphere to its translational and rotational kinetic energy when it reaches the bottom.

The formula \( v = \sqrt{\frac{10gL\sin(\theta)}{7}} \) helps calculate this translational speed, reflecting how potential energy is completely transformed into kinetic energy. The length of the incline \( L \) and the angle \( \theta \) dictate how quickly the sphere moves. Translational motion allows the sphere to cover distance on the incline, and is an essential component of analyzing physical dynamics.
Kinetic Energy Ratio
Understanding the ratio of translational to rotational kinetic energy is critical in observing how energy is distributed in the sphere as it rolls.

Translational kinetic energy \( K_{trans} \) describes the energy due to its translational speed, given by \( \frac{1}{2}mv^2 \). In contrast, rotational kinetic energy \( K_{rot} \) refers to the energy from its spin, expressed by \( \frac{1}{2}I\omega^2 \).

When computed, both forms of energy reveal crucial insights into the motion dynamics. Indeed, the ratio \( \frac{K_{trans}}{K_{rot}} \) is calculated as approximately 2.52, indicating that the translational kinetic energy of the sphere is more than double its rotational kinetic energy.

This ratio is particularly fascinating because it highlights how the energy behavior can differ even when both types have their origins from the same initial potential energy. Furthermore, calculations ensure that this ratio is independent of the sphere's radius and mass, making it universally applicable to similar scenarios.

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Most popular questions from this chapter

(II) A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 cm and accelerates at the rate of \(7.2 rad/s^2\), and it is in contact with the pottery wheel (radius 27.0 cm) without slipping. Calculate \((a)\) the angular acceleration of the pottery wheel, and \((b)\) the time it takes the pottery wheel to reach its required speed of 65 rpm.

A cyclist accelerates from rest at a rate of \(1.00 m/s^2\). How fast will a point at the top of the rim of the tire \((diameter = 68.0 cm)\) be moving after 2.25 s? [\(Hint\): At any moment, the lowest point on the tire is in contact with the ground and is at rest-see Fig. 8-57.

Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 9.0 days. If it were to undergo gravitational collapse to a neutron star of radius 12 km, losing \\(\frac{3}{4}\\) of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times. Assume also that the thrownoff mass carries off either \((a)\) no angular momentum, or \((b)\) its proportional share \((\frac{3}{4})\) of the initial angular momentum.

(II) A potter is shaping a bowl on a potter's wheel rotating at constant angular velocity of 1.6 rev/s (Fig. 8-48). The friction force between her hands and the clay is 1.5 N total. \((a)\) How large is her torque on the wheel, if the diameter of the bowl is 9.0 cm? \((b)\) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hands? The moment of inertia of the wheel and the bowl is \(0.11 kg\cdot m^2\).

A small mass \(m\) attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table (Fig. 8-62). Initially, the mass revolves with a speed \(\upsilon_1 = 2.4 m/s\) in a circle of radius \(r_1 = 0.80 m\). The string is then pulled slowly through the hole so that the radius is reduced to \(r_2 = 0.48 m\). What is the speed, \(\upsilon_2\) , of the mass now?

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