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Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length 0.600 m at the rate of 8.00 rev/s. If he increased the length to 0.900 m, he could revolve the sling only 6.00 times per second. (a) Which rate of rotation gives the greater speed for the stone at the end of the sling? (b) What is the centripetal acceleration of the stone at 8.00 rev/s? (c) What is the centripetal acceleration at 6.00 rev/s?

Short Answer

Expert verified
Greater speed is achieved with the 0.600 m sling at 8.00 rev/s. The centripetal acceleration at 8.00 rev/s is 302.4 m/s^2. At 6.00 rev/s, the centripetal acceleration is 302.4 m/s^2.

Step by step solution

01

Calculate the linear speed for the 0.600 m sling

First, convert the rotational speed from revolutions per second to radians per second by multiplying by 2π, because there are 2π radians in one revolution. The formula for linear speed (v) is v = ωr, where ω is the angular speed in radians per second and r is the radius of the circle (length of the sling). Here, ω = 8 rev/s * 2π rad/rev.
02

Calculate the linear speed for the 0.900 m sling

Using the same formula as in Step 1, calculate the linear speed for the sling with a length of 0.900 m. The angular speed here is 6 rev/s, so ω = 6 rev/s * 2π rad/rev.
03

Determine which speed is greater

Compare the two linear speeds calculated in Steps 1 and 2 to find out which is greater. The rate of rotation that gives the greater linear speed is the one with the greater value of v.
04

Calculate the centripetal acceleration for 8.00 rev/s

The formula for centripetal acceleration (a_c) is a_c = ω^2r, where ω is the angular speed and r is the radius. Use the angular speed from Step 1 for the 0.600 m sling at 8.00 rev/s.
05

Calculate the centripetal acceleration for 6.00 rev/s

Repeat the calculation of centripetal acceleration for the 0.900 m sling at 6.00 rev/s, using the angular speed from Step 2.

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Key Concepts

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

Rotational Motion
Rotational motion is a form of movement where an object spins around a central point or axis. This kind of motion is typical in everyday life, from the rotation of a bicycle wheel to the Earth spinning on its axis. In rotational motion, every point of the object moves in a circle around the axis. An important feature of this motion is that all points move at the same angular velocity, which is a measure of how fast the angle changes.

Imagine holding a string with a ball at the end and spinning it around; this is a simple model of rotational motion. The string's length represents the radius of the circular path, and the ball's path is the circumference. If you increase the string's length, the radius and, consequently, the circumference the ball travels also increase. When analyzing problems in rotational motion, two key aspects we focus on are angular speed and the force that keeps the object moving in a circular path, known as centripetal force.
Angular Speed
Angular speed is the rate at which an object rotates or revolves around an axis. It is measured in radians per second in the SI unit system. A radian is a way of measuring angles based on the radius of a circle. Specifically, one radian is the angle made at the center of a circle by an arc whose length is equal to the radius of the circle.

To understand this better, think about a pizza being sliced into equal pieces. If the length of the crust of one slice is the same as the radius of the whole pizza, that slice's tip forms an angle of one radian at the center of the pizza. To convert a rotational speed from revolutions per second to radians per second, you multiply by the constant \(2\pi\), because there are \(2\pi\) radians in one full revolution, or 360 degrees. For example, if David can rotate his sling 8 times per second, we can express this angular speed as \(8 \times 2\pi\) rad/s, which would be needed for any formulas involving angular speed, like those for linear speed and centripetal acceleration.
Linear Speed
Linear speed refers to how fast an object is moving along a straight path. In the context of rotational motion, we often talk about the 'tangential' linear speed, which is the speed of an object moving along the edge of the circle it's rotating around. It's what you experience when you're in a car going around a sharp turn: your body gets pushed to the side, but the car is still moving forward around the bend.

The linear speed can be calculated from the angular speed using the formula \( v = \omega r \), where \( v \) is the linear speed, \( \omega \) is the angular speed in radians per second, and \( r \) is the radius of the circular path. David's stone at the end of his sling is a great example. When he whirls the sling, the stone travels at a certain linear speed around a circular path. The longer the sling, the larger the radius and, therefore, the faster the stone has to travel to complete one rotation in the same amount of time.

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Most popular questions from this chapter

At \(t=0,\) a particle moving in the \(x y\) plane with constant acceleration has a velocity of \(\mathbf{v}_{i}=(3.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) and is at the origin. At \(t=3.00\) s, the particle's velocity is \(\mathbf{v}=(9.00 \hat{\mathbf{i}}+7.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) . Find \((\mathrm{a})\) the acceleration of the particle and \((\mathrm{b})\) its coordinates at any time \(t .\)

A car travels due east with a speed of 50.0 \(\mathrm{km} / \mathrm{h}\) . Rain- drops are falling at a constant speed vertically with respect to the Earth. The traces of the rain on the side windows of \(f\) the car make an angle of \(60.0^{\circ}\) with the vertical. Find the velocity of the rain with respect to (a) the car and (b) the Earth.

A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, to form a meter-high railing around the play-ground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of \(53.0^{\circ}\) above the horizontal at a point 24.0 meters from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the distance from the wall to the point on the roof where the ball lands.

As their booster rockets separate, Space Shuttle astronauts typically feel accelerations up to \(3 g,\) where \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) . In their training, astronauts ride in a device where they experience such an acceleration as a centripetal acceleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm that then turns at constant speed in a horizontal circle. Determine the rotation rate, in revolutions per second, required to give an astronaut a centripetal acceleration of 3.00g while in circular motion with radius 9.45 m.

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by the following expressions: $$\begin{array}{c}{x=(18.0 \mathrm{m} / \mathrm{s}) t} \\ {\text { and } \quad y=(4.00 \mathrm{m} / \mathrm{s}) t-\left(4.90 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}}\end{array}$$ (a) Write a vector expression for the ball's position as a function of time, using the unit vectors \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) . By taking derivatives, obtain expressions for (b) the velocity vector v as a function of time and (c) the acceleration vector a as a function of time. Next use unit- vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the golf ball, all at t " 3.00 s.

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