Question

A wheel with \(c=\frac{4}{9},\) a mass of \(40.0 \mathrm{~kg},\) and a rim radius of \(30.0 \mathrm{~cm}\) is mounted vertically on a horizontal axis. A 2.00 -kg mass is suspended from the wheel by a rope wound around the rim. Find the angular acceleration of the wheel when the mass is released.

Short answer

Answer: The angular acceleration of the wheel when the mass is released is 3.68 rad/s² (rounded to two decimal places).

Step-by-step solution

Moment of inertia

The moment of inertia of the wheel is given by \(I = c \times M \times R^2\), where \(I\) is the moment of inertia, \(c\) is a dimensionless constant, \(M\) is the mass of the wheel, and \(R\) is the radius of the wheel. First, we need to convert the radius of the wheel from centimeters to meters: \(R = \frac{30.0 \mathrm {~cm}}{100 {~cm/m}} = 0.30 \mathrm {~m}\). Now, we can find the moment of inertia: \(I = \frac{4}{9} \times 40.0 \mathrm {~kg} \times (0.30 \mathrm {~m})^2 = 1.6 \mathrm {~kg \cdot m^2}\). #Step 2: Find the torque acting on the wheel#

Torque

The torque acting on the wheel is caused by the hanging mass and can be found using the formula \(\tau = r \times F\), where \(\tau\) is the torque, \(r\) is the distance from the axis to the point on the rim where the force is applied (equal to the radius of the wheel, \(R\)), and \(F\) is the force acting on the wheel due to the hanging mass, which is equal to the gravitational force acting on the mass: \(F = m \times g\), where \(m\) is the mass of the hanging object and \(g\) is the acceleration due to gravity (\(9.81 \mathrm {~m/s^2}\)). So, the torque acting on the wheel is: \(\tau = R \times m \times g = 0.30 \mathrm {~m} \times 2.00 \mathrm {~kg} \times 9.81 \mathrm {~m/s^2} = 5.886 \mathrm {~N \cdot m}\). #Step 3: Find the angular acceleration of the wheel#

Angular acceleration

Now that we know the torque acting on the wheel and its moment of inertia, we can use Newton's second law for rotation to find the angular acceleration of the wheel. The law states that the angular acceleration \(\alpha\) is given by the torque divided by the moment of inertia: \(\alpha = \frac{\tau}{I}\). So, the angular acceleration of the wheel is: \(\alpha = \frac{5.886 \mathrm {~N \cdot m}}{1.6 \mathrm {~kg \cdot m^2}} = 3.67875 \mathrm {~rad/s^2}\). Therefore, the angular acceleration of the wheel when the mass is released is \( 3.68 \mathrm {~rad/s^2}\) (rounded to two decimal places).

Key concepts

Moment of Inertia

When delving into rotational motion, 'moment of inertia' is a crucial concept that often perplexes students. It is analogous to mass in linear motion, describing an object's resistance to changes in its rotational state. The formula commonly used to calculate this value is:
\( I = c \times M \times R^2 \), where:\
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• \(I\) represents the moment of inertia,\
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• \(c\) is a dimensionless constant unique to each object's geometry,\
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• \(M\) is the mass of the object, and\
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• \(R\) is the distance from the axis of rotation to the point of the object's mass distribution.\
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For a wheel, this constant \(c\) differentiates diverse shapes, reflecting the distribution of mass around the wheel's axis. In this exercise, with \(c = \frac{4}{9}, M = 40 \mathrm{~kg}\), and \(R = 0.30 \mathrm{~m}\), we deduce the wheel's moment of inertia to be \(1.6 \mathrm{~kg \cdot m^2}\r). Understanding moment of inertia is vital since it is directly proportional to the torque required to achieve a certain angular acceleration.

Torque

Imagine trying to open a heavy door; the effort you apply on the door handle is akin to 'torque' in the realm of physics. Torque, in essence, is the measure of the force causing an object to rotate around an axis. It's calculated using the formula:
\( \tau = r \times F \), where:\
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• \(\tau\) is the torque exerted on the object,\
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• \(r\) is the lever arm, the radial distance from the pivot point to the point where force is applied, and\
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• \(F\) is the force being exerted perpendicular to the lever arm.\
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In our textbook exercise, the hanging mass exerts a force due to gravity (its weight), causing a torque on the wheel. The force calculates as \(F = m \times g\r), where \(g = 9.81 \mathrm{~m/s^2}\r) is the gravitational acceleration. With \(m = 2.00 \mathrm{~kg}\) and \(r = R = 0.30 \mathrm{~m}\), the torque comes out to be \(5.886 \mathrm{~N \cdot m}\). Knowing this force-arm product helps us to analyze and predict the wheel's rotational motion.

Newton's Second Law for Rotation

Newton's second law for rotation is a foundational concept that extends the familiar force-equals-mass-times-acceleration (\( F=ma \)) from linear to rotary motion. It states that the angular acceleration (\(\alpha\r)) of an object is directly proportional to the net external torque (\(\tau\r)) and inversely proportional to the object's moment of inertia (\(I\r)). This relationship is codified in the equation:
\( \alpha = \frac{\tau}{I} \).

For our vertical wheel exercise, after calculating the torque (\(5.886 \mathrm{~N \cdot m}\)) and knowing the moment of inertia (\(1.6 \mathrm{~kg \cdot m^2}\)), we can determine the wheel's angular acceleration. Plug these figures into the law, and we obtain \(\alpha = 3.67875 \mathrm{~rad/s^2}\r), or rounded, \(3.68 \mathrm{~rad/s^2}\r). Understanding this relationship is paramount for predicting how quickly an object will start to spin or stop spinning when a torque is applied, which has practical applications in everything from engineering to everyday gadgets.