4.4 Uniform circular motion (Page 4/8)

University physics volume 1 Page 4 / 8

Total acceleration during circular motion

A particle moves in a circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varies with time according to

v (t) = c_{1} - \frac{c_{2}}{t^{2}}, c_{1} = 4.0 m / s, c_{2} = 6.0 m \cdot s .

What is the total acceleration of the particle at t = 2.0 s?

Strategy

We are given the speed of the particle and the radius of the circle, so we can calculate centripetal acceleration easily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of the tangential acceleration by taking the derivative with respect to time of

| v (t) |

using [link] and evaluating it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration.

Solution

Centripetal acceleration is

v (2.0 s) = (4.0 - \frac{6.0}{{(2.0)}^{2}}) m / s = 2.5 m / s

a_{C} = \frac{v^{2}}{r} = \frac{(2.5 m / {s)}^{2}}{2.0 m} = 3.1 m / s^{2}

directed toward the center of the circle. Tangential acceleration is

a_{T} = | \frac{d \vec{v}}{d t} | = \frac{2 c_{2}}{t^{3}} = \frac{12.0}{{(2.0)}^{3}} m / s^{2} = 1.5 m / s^{2} .

Total acceleration is

| \vec{a} | = \sqrt{{3.1}^{2} + {1.5}^{2}} m / s^{2} = 3.44 m / s^{2}

and $θ = {tan}^{−1} \frac{3.1}{1.5} = 64 °$ from the tangent to the circle. See [link] .

The acceleration of a particle on a circle is shown along with its radial and tangential components. The centripetal acceleration a sub c points radially toward the center of the circle and has magnitude 3.1 meters per second squared. The tangential acceleration a sub T is tangential to the circle at the particle’s position and has magnitude 1.5 meters per second squared. The angle between the total acceleration a and the tangential acceleration a sub T is 64 degrees. — The tangential and centripetal acceleration vectors. The net acceleration $\vec{a}$ is the vector sum of the two accelerations.

Significance

The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polar coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used for motion along curved paths, is discussed in detail later in the book.

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Summary

Uniform circular motion is motion in a circle at constant speed.
Centripetal acceleration ${\vec{a}}_{C}$ is the acceleration a particle must have to follow a circular path. Centripetal acceleration always points toward the center of rotation and has magnitude $a_{C} = v^{2} / r .$
Nonuniform circular motion occurs when there is tangential acceleration of an object executing circular motion such that the speed of the object is changing. This acceleration is called tangential acceleration ${\vec{a}}_{T} .$ The magnitude of tangential acceleration is the time rate of change of the magnitude of the velocity. The tangential acceleration vector is tangential to the circle, whereas the centripetal acceleration vector points radially inward toward the center of the circle. The total acceleration is the vector sum of tangential and centripetal accelerations.
An object executing uniform circular motion can be described with equations of motion. The position vector of the object is $\vec{r} (t) = A cos ω t \hat{i} + A sin ω t \hat{j},$ where A is the magnitude $| \vec{r} (t) |,$ which is also the radius of the circle, and $ω$ is the angular frequency.

Conceptual questions

Can centripetal acceleration change the speed of a particle undergoing circular motion?

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Can tangential acceleration change the speed of a particle undergoing circular motion?

yes

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Problems

A flywheel is rotating at 30 rev/s. What is the total angle, in radians, through which a point on the flywheel rotates in 40 s?

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A particle travels in a circle of radius 10 m at a constant speed of 20 m/s. What is the magnitude of the acceleration?

$a_{C} = 40 m / s^{2}$

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Cam Newton of the Carolina Panthers throws a perfect football spiral at 8.0 rev/s. The radius of a pro football is 8.5 cm at the middle of the short side. What is the centripetal acceleration of the laces on the football?

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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00-m radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity?

$a_{C} = \frac{v^{2}}{r} \Rightarrow v^{2} = r a_{C} = 78.4, v = 8.85 m / s$
$T = 5.68 s,$ which is $0.176 rev / s = 10.6 rev / min$

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A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of 30.0 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 23.2 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?

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What is the acceleration of Venus toward the Sun, assuming a circular orbit?

Venus is 108.2 million km from the Sun and has an orbital period of 0.6152 y.
$r = 1.082 \times 10^{11} m T = 1.94 \times 10^{7} s$
$v = 3.5 \times 10^{4} m/s, a_{C} = 1.135 \times 10^{−2} {m/s}^{2}$

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An experimental jet rocket travels around Earth along its equator just above its surface. At what speed must the jet travel if the magnitude of its acceleration is g ?

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A fan is rotating at a constant 360.0 rev/min. What is the magnitude of the acceleration of a point on one of its blades 10.0 cm from the axis of rotation?

$360 rev / min = 6 rev / s$
$v = 3.8 m / s$ $a_{C} = 144. m / s^{2}$

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A point located on the second hand of a large clock has a radial acceleration of $0.1 cm / s^{2} .$ How far is the point from the axis of rotation of the second hand?

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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