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Magazine Chapter 9: Problem 29
A flywheel with radius \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2} .\) For a point on the rim of the flywheel, what are the magnitudes of the tangential, radial, and resultant accelerations after \(2.00 \mathrm{~s}\) of acceleration?
Short Answer
Expert verified
The magnitudes of the tangential, radial, and resultant accelerations after 2.00 s of acceleration are 0.180 m/s^2, 0.432 m/s^2, and 0.465 m/s^2, respectively.
Step by step solution
01
Determine Tangential Acceleration
The tangential acceleration (tangential component of linear acceleration) of a point on the rim of the flywheel after a certain time can be found using the formula \(a_t = \alpha r\) where \(a_t\) is the tangential acceleration, \(\alpha\) is the angular acceleration, and \(r\) is the radius of the wheel. Plugging in the given values we get \(a_t = 0.600 \, rad/s^2 \times 0.300 \, m = 0.180 \, m/s^2\).
02
Calculate Radial Acceleration
The radial acceleration (centripetal component of linear acceleration) at any instance in a rotating system like this can be calculated using the formula \(a_r = 蠅虏r\). However, the angular speed 蠅 is not given here, so you have to use another formula: \(蠅 = 伪t\) to calculate it first, where \(t\) is the time that the system has been accelerating. Plugging in the given values we get \(蠅 = 0.600 \, rad/s^2 \times 2.00 \, s = 1.20 \, rad/s\). Then we can substitute \( 蠅 \) back into \(a_r = 蠅虏r\): \(a_r = (1.20 \, rad/s)^2 \times 0.300 \, m = 0.432 \, m/s^2 \).
03
Find Resultant Acceleration
The resultant acceleration (as combined effect of both tangential and radial acceleration) can be worked out by the formula \( a = \sqrt{a_t^2 + a_r^2}\). Plugging in our values we get \(a = \sqrt{(0.180 \, m/s^2)^2 + (0.432 \, m/s^2)^2} = 0.465 \, m/s^2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangential Acceleration
When an object rotates, such as a point on the rim of a flywheel, it experiences different forms of acceleration. One of these is tangential acceleration, which is the rate at which the linear (or tangential) speed of the object changes. It represents how quickly the velocity of the object is changing along the edge of the circular path.
For an object moving in circular motion with a constant angular acceleration, the tangential acceleration is a linear acceleration tangent to the circle at the object's position. It can easily be calculated using the formula:
\[\begin{equation} a_t = \text{angular acceleration} \times r \end{equation}\]where \(a_t\) is the tangential acceleration, and \(r\) is the radius of the object's circular path. The SI unit for tangential acceleration is meters per second squared \(m/s^2\). An important point to remember is that tangential acceleration only affects the magnitude of velocity, not its direction.
In our flywheel example, the angular acceleration provided is \(0.600 \text{rad/s}^2\), and the flywheel's radius is \(0.300 \text{m}\). Therefore:
\[\begin{equation}a_t = 0.600 \text{rad/s}^2 \times 0.300 \text{m} = 0.180 \text{m/s}^2\end{equation}\]This is how we measure the change in speed for a point on the flywheel's rim over time.
For an object moving in circular motion with a constant angular acceleration, the tangential acceleration is a linear acceleration tangent to the circle at the object's position. It can easily be calculated using the formula:
\[\begin{equation} a_t = \text{angular acceleration} \times r \end{equation}\]where \(a_t\) is the tangential acceleration, and \(r\) is the radius of the object's circular path. The SI unit for tangential acceleration is meters per second squared \(m/s^2\). An important point to remember is that tangential acceleration only affects the magnitude of velocity, not its direction.
In our flywheel example, the angular acceleration provided is \(0.600 \text{rad/s}^2\), and the flywheel's radius is \(0.300 \text{m}\). Therefore:
\[\begin{equation}a_t = 0.600 \text{rad/s}^2 \times 0.300 \text{m} = 0.180 \text{m/s}^2\end{equation}\]This is how we measure the change in speed for a point on the flywheel's rim over time.
Radial Acceleration
Radial acceleration, on the other hand, refers to the object's acceleration towards the center of its circular path, often called centripetal acceleration. This force is what keeps the object moving in a circle rather than flying off in a straight line as per Newton's first law of motion.
The formula for radial acceleration is given by: \[\begin{equation} a_r = \text{angular velocity}^2 \times r \end{equation}\]where \(a_r\) is the radial acceleration, and the angular velocity \(蠅\) is normally in radians per second \(rad/s\). The radius \(r\) is the same as used in the tangential acceleration formula. However, if the angular velocity is not provided, we can first calculate it using the angular acceleration \(伪\) and the time \(t\) with the equation \(蠅 = 伪t\).
In our example, after calculating \(蠅 = 0.600 \text{rad/s}^2 \times 2.00 \text{s} = 1.20 \text{rad/s}\), we use it to find the radial acceleration:\[\begin{equation}a_r = (1.20 \text{rad/s})^2 \times 0.300 \text{m} = 0.432 \text{m/s}^2\end{equation}\]This acceleration is directed towards the center of the flywheel and is responsible for changing the direction of the point's velocity vector.
The formula for radial acceleration is given by: \[\begin{equation} a_r = \text{angular velocity}^2 \times r \end{equation}\]where \(a_r\) is the radial acceleration, and the angular velocity \(蠅\) is normally in radians per second \(rad/s\). The radius \(r\) is the same as used in the tangential acceleration formula. However, if the angular velocity is not provided, we can first calculate it using the angular acceleration \(伪\) and the time \(t\) with the equation \(蠅 = 伪t\).
In our example, after calculating \(蠅 = 0.600 \text{rad/s}^2 \times 2.00 \text{s} = 1.20 \text{rad/s}\), we use it to find the radial acceleration:\[\begin{equation}a_r = (1.20 \text{rad/s})^2 \times 0.300 \text{m} = 0.432 \text{m/s}^2\end{equation}\]This acceleration is directed towards the center of the flywheel and is responsible for changing the direction of the point's velocity vector.
Resultant Acceleration
When discussing circular motion, it's important to understand that the resultant acceleration is the vector sum of tangential and radial accelerations. This combined acceleration illustrates the overall effect of both changing speed and changing direction on a point moving in a circle.
Given that the tangential and radial accelerations are at right angles to each other (tangent and radius of a circle always meet at a right angle), the resultant acceleration can be calculated using the Pythagorean theorem:\[\begin{equation} a = \text{resultant acceleration} = \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ 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\text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\text: 魔\text: K\text: 脨\text: Q\text: 脨\text: 脨\text: Z\text: K\text: P\text: M\text: 脨\text: P\text: 脨\text: 1\text: 4\text: 0\text: 0\text: 7\text: 0\text: 4\text: 2\text: 1\text: 0\text: 5\text: 1\text: 2\text: X\text: Y\text: 膭\text: T\text: 臉\text: 膯\text: 呕\text: 膭\text: 殴\text: 艂\text: 艧\text: 臒\text: 5\text: 3\text: 6\text: 4\text: 8\text: 4\text: 3\text: 2\text: 5\text: 4\text: 4\text: 8\text: 3\text: 7\text: 4\text: 8\text: 2\text: 4\text: 1\text: 3\text: 1\text: 6: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\end{equation}\]Once the tangential acceleration \(a_t\) and radial acceleration \(a_r\) are known, use their values to find the resultant acceleration:\[\begin{equation}a = \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\end{equation}\]In the flywheel example, we calculate it using the previous findings:\[\begin{equation} a = \text{ }\text{ }\text: \text: \text: 魔\end{equation}\]This \(0.465 \text{m/s}^2\) is the overall linear acceleration experienced by a point on the rim of the flywheel after 2 seconds.
Given that the tangential and radial accelerations are at right angles to each other (tangent and radius of a circle always meet at a right angle), the resultant acceleration can be calculated using the Pythagorean theorem:\[\begin{equation} a = \text{resultant acceleration} = \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ 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\text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\text: 魔\text: K\text: 脨\text: Q\text: 脨\text: 脨\text: Z\text: K\text: P\text: M\text: 脨\text: P\text: 脨\text: 1\text: 4\text: 0\text: 0\text: 7\text: 0\text: 4\text: 2\text: 1\text: 0\text: 5\text: 1\text: 2\text: X\text: Y\text: 膭\text: T\text: 臉\text: 膯\text: 呕\text: 膭\text: 殴\text: 艂\text: 艧\text: 臒\text: 5\text: 3\text: 6\text: 4\text: 8\text: 4\text: 3\text: 2\text: 5\text: 4\text: 4\text: 8\text: 3\text: 7\text: 4\text: 8\text: 2\text: 4\text: 1\text: 3\text: 1\text: 6: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\end{equation}\]Once the tangential acceleration \(a_t\) and radial acceleration \(a_r\) are known, use their values to find the resultant acceleration:\[\begin{equation}a = \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text{ }\text{ }\text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: \text: 魔\end{equation}\]In the flywheel example, we calculate it using the previous findings:\[\begin{equation} a = \text{ }\text{ }\text: \text: \text: 魔\end{equation}\]This \(0.465 \text{m/s}^2\) is the overall linear acceleration experienced by a point on the rim of the flywheel after 2 seconds.
