Measuring I. As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.640 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope, as in Fig. E9.45. You release the system from rest and findthat the block descends 12.0 m in 4.00 s. What is the moment of

inertia of the wheel for this axis?

Given in the question,

Diameter of the wheel,\(d = \,0.640\,\,{\rm{m}}\)

Mass of the wood block,\(m = \,8.20\,\,{\rm{kg}}\)

Height covered by the block,\(h = \,12.\,0\,\,{\rm{m}}\)

Time to cover the height,\(t = 4.00\,\,{\rm{s}}\)

Law of conservation of energy

The initial total energy is always equal to the final total energy of the system.

From the conservation of energy, we know

Total initial energy = Total final energy

Initial potential energy +initial kinetic energy= final potential energy + final kinetic energy.

\({U_i} + K{E_i} = {U_f} + K{E_f}\)…(i)

As we know initial the block is at height h and it is at the rest

therefore

the initial potential energy\({U_1} = mgh\)

where m is mass and h is the height

initial kinetic energy\(K{E_i} = 0\)

Since the final height of the block is zero

Therefore,

Final potential energy\({U_f} = 0\)

The final kinetic energy will be the sum of translational kinetic energy and rotational kinetic energy

Final kinetic energy

\(\begin{aligned}{}K{E_f} = K{E_t} + K{E_r}\\ = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\end{aligned}\)

Where I is the moment of inertia and\(\omega \)is angular velocity.

Substituting all the values into the equation (i)

\(mgh + 0 = 0 + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\)

We know,

\(v = r\omega \)

\(mgh = \frac{1}{2}m{v^2} + \frac{1}{2}I{\left( {\frac{v}{r}} \right)^2}\)….(ii)

To find the velocity using the equation.

\(v = u + at\)…(iii)

\(h = ut + \frac{1}{2}a{t^2}\)…(iv)

Where u is initial velocity, v is final velocity, a is acceleration

And t is time.

We know

\(\begin{aligned}{}u = 0\\t = \,4\,\,{\rm{s}}\end{aligned}\)

\(h = \,12\,\,{\rm{m}}\)

Substituting the values into the equations (iii)

\(v = \left( {4\,{\rm{s}}} \right)a\)….(v)

Substituting the values into equation (iv)

\(\begin{aligned}{}12\,\,{\rm{m}} = 0\left( {4.00\,{\rm{s}}} \right) + \frac{1}{2}a{(4.00\,\,{\rm{s}})^2}\\a = \,1.5\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

Therefore, from equation (v)

\(\begin{aligned}{}v = \left( {4\,{\rm{s}}} \right)\left( {1.5\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\\v = 6\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.}{\rm{s}}}\end{aligned}\)

Now, from equation (i)

\(mgh = \frac{1}{2}m{v^2} + \frac{1}{2}I{\left( {\frac{v}{r}} \right)^2}\)

Substituting the values

\(\begin{aligned}{}\left( {\,8.20\,\,{\rm{kg}}} \right)\left( {9.8\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\,\left( {12.\,0\,\,{\rm{m}}} \right) = \frac{1}{2}\left( {\,8.20\,\,{\rm{kg}}} \right){\left( {6\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2} + \frac{1}{2}I{\left( {\frac{{6\,\,{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{\,{{0.640\,\,{\rm{m}}} \mathord{\left/{\vphantom {{0.640\,\,{\rm{m}}} 2}} \right.} 2}}}} \right)^2}\\\frac{1}{2}I{\left( {\frac{{6\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{\,{{0.640\,\,{\rm{m}}} \mathord{\left/{\vphantom {{0.640\,\,{\rm{m}}} 2}} \right.} 2}}}} \right)^2} = \left( {\,8.20\,\,{\rm{kg}}} \right)\left( {9.8\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\,\left( {12.\,0\,\,{\rm{m}}} \right) - \frac{1}{2}\left( {\,8.20\,\,{\rm{kg}}} \right){\left( {6\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2}\\I = \frac{{2\left( {\,8.20\,\,{\rm{kg}}} \right)\left( {9.8\,\,{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\,\left( {12.\,0\,\,{\rm{m}}} \right) - \left( {\,8.20\,\,{\rm{kg}}} \right){{\left( {6\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}^2}}}{{{{\left( {\frac{{6\,\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}}}{{\,{{0.640\,\,{\rm{m}}} \mathord{\left/{\vphantom {{0.640\,\,{\rm{m}}} 2}} \right.} 2}}}} \right)}^2}}}\\I = \frac{{167.33\,\,}}{{36}}{\rm{kg - }}{{\rm{m}}^{\rm{2}}}\\{\rm{So}}\\I = 4.65\,\,{\rm{kg - }}{{\rm{m}}^{\rm{2}}}\end{aligned}\)

Hence the moment of inertia of the wheel is \(I = 4.65\,\,{\rm{kg - }}{{\rm{m}}^{\rm{2}}}\)