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Chapter 10: Problem 51

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

Expert verified
#tag_title#Answer#tag_content# When the mass is released, the angular acceleration of the wheel is given by the formula: \(α = \frac{\tau}{I} = \frac{rF}{\frac{1}{2}MR^2}\). First, we calculate the force exerted by the suspended mass: \(F = mg = (2.00 \mathrm{~kg})(9.81 \mathrm{~m/s^2}) = 19.62 \mathrm{~N}\). Next, we calculate the moment of inertia of the wheel: \(I = \frac{1}{2}(40.0 \mathrm{~kg})(0.3 \mathrm{~m})^2 = 1.8 \mathrm{~kg \cdot m^2}\). Now, we substitute the values of torque and moment of inertia into the formula for angular acceleration: \(α = \frac{(0.3 \mathrm{~m})(19.62 \mathrm{~N})}{1.8 \mathrm{~kg \cdot m^2}} = 3.27 \mathrm{~rad/s^2}\). Therefore, the angular acceleration of the wheel when the mass is released is \(3.27 \mathrm{~rad/s^2}\).

Step by step solution

01

Identify the forces acting on the system

There are two forces acting on the system: the gravitational force acting on the suspended mass and the tension in the rope. The gravitational force acting on the mass causes it to accelerate downward and exerts a torque on the wheel, causing it to rotate. The tension in the rope opposes the gravitational force and also exerts a torque on the wheel.
02

Determine the moment of inertia of the wheel

To calculate the moment of inertia for the wheel, we can use the formula for a solid cylinder, \(I = \frac{1}{2}MR^2\) where \(M\) is the mass of the wheel, and \(R\) is the rim radius. The moment of inertia of the wheel is: \(I = \frac{1}{2}(40.0 \mathrm{~kg})(0.3 \mathrm{~m})^2\)
03

Calculate the torque acting on the wheel due to the suspended mass

The torque acting on the wheel due to the suspended mass can be estimated using the formula: \(\tau = rF\), where \(r\) is the radius of the rim, and \(F\) is the force exerted by the suspended mass (the gravitational force acting on the mass). The force exerted by the suspended mass is: \(F = mg\), where \(m = 2.00 \mathrm{~kg}\) (the mass of the suspended weight), and \(g = 9.81 \mathrm{~m/s^2}\) (acceleration due to gravity).
04

Use Newton's second law for rotation to find the angular acceleration

Newton's second law for rotation states that the angular acceleration (α) is equal to the torque (τ) divided by the moment of inertia (I): \(α = \frac{\tau}{I}\). Substituting the expressions for torque and moment of inertia, we find the angular acceleration when the mass is released.

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