A thin, light wire is wrapped around the rim of a wheelas shown in Fig. E9.45. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has a radius 0.180 m and a moment of inertia for rotation about the axle of $\mathit{I}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{480}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$. A small block with a mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does -9.00 J of work as the block descends 3.00 m. What is themagnitude of the angular velocity of the wheel after the block hasdescended 3.00 m?

Given in the equation

The radius of the wheel, r=0.180 m

The moment of inertia of the pulley, l=0.480 kg m²

The mass of the block m=0.340 kg

Work done by friction, W_f=-9.00 J

Distance covered by the block d=3m

Law of conservation of energy

According to the conservation of energy, the energy of a systemcan neither be created nor destroyed it can only be converted from one form of energy to another.

$\begin{array}{l}{E}_F=E_I\\ U_i+K{E}_i+W_{other}=U_f+K{E}_f\end{array}$

Where,

$U_i$ is initial potential energy, $U_f$is final potential energy, $K{E}_i$is initial kinetic energy, $K{E}_f$ is final kinetic energy, and $W_{other}$is work done by other forces.

From the conservation of energy,

$U_i+K{E}_i+W_{other}=U_f+K{E}_f$…(i)

Since the wire does not have any mass its potential and kinetic energy are always zero.

potential energy can be given as

$U=mgh$

Where Uis potential energy, m is mass and h is the height.

$\begin{array}{l}{U}_i=mgd\\ U_i=\left(0.340\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\left(9.8\text{\hspace{0.17em}\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(3m\right)\\ U_i=9.996\text{\hspace{0.17em}J}\end{array}$

Since finally the height of the block is zero

Final potential energy $U_f=0$

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

The formula of transitional kinetic energy is

$K_t=\frac{1}{2}m{v}^2$

Where m is mass and v is the linear speed

The formula of rotational kinetic energy

$k_r=\frac{1}{2}I{\omega }^2$

Where I is inertia and is the angular velocity

Initially, the system was at rest therefore initial kinetic energy.

$K{E}_i=0$

Final kinetic energy

$\begin{array}{c}K{E}_f=K{E}_{tB}+K{E}_{rw}\\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\end{array}$

We know, the block and wheel are connected by wire so they will move with the same linear speed

Therefore, we can use the formula

$v=r\omega$

Substituting the values

$\begin{array}{c}K{E}_f=K{E}_{tB}+K{E}_{rw}\\ =\frac{1}{2}m{\left(r\omega \right)}^2+\frac{1}{2}I{\omega }^2\\ =\left[\frac{1}{2}m{\left(r\right)}^2+\frac{1}{2}I\right]{\omega }^2\\ =\left[\frac{1}{2}\left(0.340\text{\hspace{0.17em}\hspace{0.17em}kg}\right){\left(0.180\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2+\frac{1}{2}\left(0.480{\text{\hspace{0.17em}\hspace{0.17em}kg.m}}^{\text{2}}\right)\right]{\omega }^2\\ =0.2455{\omega }^2\text{\hspace{0.17em}\hspace{0.17em}J}\end{array}$

Since the only force acting on the system is friction

Therefore

$\begin{array}{c}{W}_{other}=W_{friction}\\ =-9.00\text{\hspace{0.17em}\hspace{0.17em}J}\end{array}$

Substituting all the values of energy and work in the equation (i)

$\begin{array}{l}9.996\text{\hspace{0.17em}J}+0-9.00\text{\hspace{0.17em}\hspace{0.17em}J}=0+0.2455{\omega }^2\\ \omega =2.01\text{\hspace{0.17em}\hspace{0.17em}rad}/\text{s}\end{array}$

Hence the angular speed of the wheel is $2.01rad/s$