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A fly wheel having mass $3\text{ kg}$ and radius $5\text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to $3\text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by $3\text{ m}$ is _________ $\text{J}$. ($g = 10\text{ m/s}^{2}$)

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} $
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Important Questions from Mechanics

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  5. 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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