Question

A smooth circular hoop with a radius of 0.500 m is placed flat on the floor. A particle slides around the inside edge of the hoop. The particle is given an initial speed of $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. After one revolution, its speed has dropped to$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ because of friction with the floor. (a) Find the energy transformed from mechanical to internal in the particle hoop floor system as a result of friction in one revolution. (b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

Short answer

(a)The energy transformed from mechanical to internal in the particle hoop floor system as a result of friction in one revolution is $\Delta E_{int}=5.60J$

(b) The total number of revolutions the particle makes before stopping is $N=2.28rev$.

Step-by-step solution

Concept and Formula of energy

Non isolated System (Energy): The most general statement describing the behavior of a non isolated system is the conservation of energy equation:

$\Delta E_{system}=\sum T..........\left(8.1\right)$

Including the types of energy storage and energy transfer that we have discussed gives

$\Delta K+\Delta U+\Delta E_{int}=W+Q+T_{MW}+T_{ET}+T_{ER}$

For a specific problem, this equation is generally reduced to a smaller number of terms by eliminating the terms that are not appropriate to the situation.

The change in the kinetic energy is given by

$\Delta k=\frac{1}{2}m\left({v_f}^2-{v_i}^2\right)$

where

$\Delta k=$Change in kinetic energy

$m$is mass of object

$v_f=$Final speed of the object

$v_f=$Initial speed of the object.

Calculating the energy transfer

(a)

From using the equation (8.1), we have

$\Delta E_{int}=-\Delta k$

Since we know that,$\Delta k=\frac{1}{2}m\left({v_f}^2-{v_i}^2\right)$

$\Delta E_{int}=-\frac{1}{2}m\left({v_f}^2-{v_i}^2\right)$

Substituting values, we get

$$\begin{gathered} \Delta E_{int}=-\frac{1}{2}\left(0.4kg\right)\left({\left(6m/s\right)}^2-{\left(8m/s\right)}^2\right) \\ \Delta E_{int}=5.60J \end{gathered}$$

Calculating the total number of revolutions

(b)

It is given that the friction force remains constant during the entire motion.

So, after N revolutions, the object comes to rest and$k_f=0$

Thus

$$\begin{gathered} \Delta E_{int}=-\Delta k \\ \Delta E_{int}=f_{k}d \end{gathered}$$

Therefore

$$\begin{gathered} \Delta E_{int}=-\Delta k \\ {f}_{k}d=-\left(0-k_i\right) \\ {f}_{k}d=\frac{1}{2}m{v}_i^2 \\ \left({\mu }_{k}mg\right)\left(N2\pi r\right)=\frac{1}{2}m{v}_i^2...........\left(\because f_k={\mu }_{k}mg,d=N2\pi r\right) \end{gathered}$$

This gives

$$\begin{gathered} N=\frac{\frac{1}{2}m{v}_i^2}{{\mu }_{k}mg\times 2\pi r} \\ N=\frac{\frac{1}{2}\left(0.4kg\right){\left(8.00m/s\right)}^2}{\left(0.152\right)\left(9.8m/s^2\right)2\pi \left(1.5m\right)} \\ N=2.28rev \end{gathered}$$