Question

In Figure a wheel of radius $\mathbf{0}\mathbf{.}\mathbf{20}\mathit{m}$ is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is $\mathbf{0}\mathbf{.}\mathbf{40}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$. A mass less cord wrapped around the wheel’s circumference is attached to a$\mathbf{6}\mathbf{.}\mathbf{0}\mathit{k}\mathit{g}$ box. The system is released from rest. When the box has a kinetic energy of $\mathbf{6}\mathbf{.}\mathbf{0}\mathit{j}$,

(a) The wheel’s rotational kinetic energy and

(b) The distance the box has fallen?

Short answer

1. Wheel’s rotational kinetic energy is $10\text{J}.$
2. Distance through which the box has fallen is$0.27\text{m}$

Step-by-step solution

Step 1: Given

1. Wheel radius is $0.20\text{m}$
2. Rotational inertia is$0.40\text{kg}-{\text{m}}^2$
3. Mass of box is$6.0\text{kg}$
4. Kinetic energy of box is $6.0\text{J}$

Determining the concept

Use the formula in terms of inertia and angular velocity to find the rotational kinetic energy. Then, using the law of conservation of energy, find the height through which the box has fallen.

Formulae are as follow:

1. $K=\frac{1}{2}m{v}^2$
2. $K_{rot}=\frac{1}{2}I{\omega }^2$
3. $v=r\omega$

Where,

K is kinetic energy, m is mass, r is radius, I is moment of inertia, v is velocity and is angular velocity.

(a) determining the wheel's rotational kinetic energy

First, find the velocity from kinetic energy as follows:

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ 6=\frac{1}{2}6v^2 \\ v=1.41\text{m}/\text{s} \end{gathered}$$

Now, find angular velocity as follows:

$$\begin{gathered} v=r\omega \\ 1.41=0.2\omega \\ \omega =7.05\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right. \end{gathered}$$

Now, rotational kinetic energy is as follows:

$$\begin{gathered} K{E}_{rotational}=0.5\times I{\omega }^2 \\ K{E}_{rotational}=0.5\times 0.40\times 7.{05}^2 \\ K{E}_{rotational}=9.9405\text{J} \end{gathered}$$

In one significant figure,

$K{E}_{rotational}=10\text{J}$

Hence, wheel’s rotational kinetic energy is $10\text{J}$

(b) Determining the distance through which the box has fallen

Now, conservation of energy,

$$\begin{gathered} K_i+U_i=K_f+U_f \\ {K}_i+U_i=\left(K_{translation}+K_{rotational}\right)+U_f \\ 0+0=\left(6+10\right)-6\times 9.8\times h \\ h=0.27\text{m} \end{gathered}$$

Hence, distance through which the box has fallen is$0.27\text{m}$

Therefore, using the formula for rotational kinetic energy and law of conservation of energy, the distance through which the box has fallen can be found.