The problem asks for the kinetic energy of a flywheel after a connected mass descends.
Flywheel Parameters Identification
- Mass of flywheel, $M = 3\text{ kg}$
- Radius of flywheel, $R = 5\text{ m}$
- Mass connected, $m = 3\text{ kg}$
- Distance descended, $h = 3\text{ m}$
- Acceleration due to gravity, $g = 10\text{ m/s}^2$
Energy Conservation Principle
The total potential energy lost by the descending mass is converted into the kinetic energy of the mass and the kinetic energy of the rotating flywheel. We assume the flywheel is a solid disk for calculating its moment of inertia.
The energy conservation equation is:
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $Where:
- $v$ is the linear velocity of the mass.
- $\omega$ is the angular velocity of the flywheel.
- $I$ is the moment of inertia of the flywheel.
Relationship between linear and angular velocity: $v = R\omega$ or $\omega = v/R$.
Moment of Inertia Calculation
Assuming the flywheel is a solid disk, its moment of inertia ($I$) is:
$ I = \frac{1}{2}MR^2 $ $ I = \frac{1}{2}(3\text{ kg})(5\text{ m})^2 = \frac{1}{2}(3)(25) = 37.5\text{ kg m}^2 $Solving for Velocity
Substitute $\omega = v/R$ and $I = 37.5\text{ kg m}^2$ into the energy equation:
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 $ $ mgh = \frac{1}{2}mv^2 + \frac{1}{4}MR^2\frac{v^2}{R^2} $ $ mgh = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2 $ $ mgh = \frac{1}{2}v^2\left(m + \frac{1}{2}M\right) $Plug in the values:
$ (3\text{ kg})(10\text{ m/s}^2)(3\text{ m}) = \frac{1}{2}v^2\left(3\text{ kg} + \frac{1}{2}(3\text{ kg})\right) $ $ 90\text{ J} = \frac{1}{2}v^2\left(3 + 1.5\right)\text{ kg} $ $ 90\text{ J} = \frac{1}{2}v^2(4.5\text{ kg}) $ $ 180\text{ J} = v^2(4.5\text{ kg}) $ $ v^2 = \frac{180}{4.5}\text{ m}^2/\text{s}^2 = 40\text{ m}^2/\text{s}^2 $Calculating Wheel's Kinetic Energy
The kinetic energy of the flywheel ($KE_{wheel}$) is:
$ KE_{wheel} = \frac{1}{2}I\omega^2 = \frac{1}{2}I\left(\frac{v}{R}\right)^2 = \frac{1}{2}\frac{I}{R^2}v^2 $We know $I = 37.5\text{ kg m}^2$ and $R = 5\text{ m}$, so $\frac{I}{R^2} = \frac{37.5}{25} = 1.5\text{ kg}$
$ KE_{wheel} = \frac{1}{2}(1.5\text{ kg})(40\text{ m}^2/\text{s}^2) $ $ KE_{wheel} = \frac{1}{2}(60\text{ J}) $ $ KE_{wheel} = 30\text{ J} $Similar Questions
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Important Questions from Mechanics
- A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
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A fly wheel of mass $10$ kg and radius $50$ cm is rotating at a rate of$360$ rpm. Assuming mass to be concentrated at the rim, the constant retarding torque required to stop the rotation of wheel in $6$ rotations will be x Nm. The value of x is ____________.
