Solved Problem on Circular Motion

A pulley A, of radius 0.15 m, begins its motion from the rest with a constant angular acceleration of 2 rad/s². This pulley is connected to a wheel B, of radius 0.40 m, by a belt that spins without slip. Determine the speed and acceleration of a point P over wheel B after two revolutions.

Problem Data:

Radius of pulley A: r_A = 0.15 m;
Acceleration of pulley A: α_A = 2 rad/s²;
Radius of pulley B: r_B = 0.40 m;
Angular displacement of point B: φ_B = 2 rotações
Initial speed of pulley A: v_0A = 0;
Initial angular speed of pulley A: ω_0A = 0;
Initial speed of wheel B: v_0B = 0;
Initial angular speedd of wheel B: ω_0B = 0.

Solution

First, we need to convert the displacement of point B given at rotations to radians (rad)

\[ \varphi_{B}=2\;\cancel{\text{rotações}}\times\frac{2\pi\;\text{rad}}{1\;\cancel{\text{rotação}}}=4\pi \;\text{rad} \]

The problem tells us that in the belt between the pulleys there is no slip, so the pulleys rotate together, and the displacement is the same for all points of the belt and the points over the pulley and the wheel. The displacement of a point in circular motion is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S=\varphi r} \tag{I} \end{gather} \]

Where S_A is the displacement of a point of pulley A (from 1 to 2, Figure 1), it will be equal to S_B, the displacement of a point of wheel B, using the expression (I), we can write the condition

\[ \begin{gather}S_{A}=S_{B}\\ \varphi_{A}r_{A}=\varphi_{B}r_{B}\\ \varphi_{A}=\frac{\varphi_{B}r_{B}}{r_{A}} \end{gather} \]

Figure 1

substituting the data of the problem and assuming π = 3.14, we find the angular displacement of a point in the pulley A

\[ \begin{gather} \varphi_{A}=\frac{4\times3.14\times 0.40}{0.15}\\ \varphi_{A}=33.5\;\text{rad} \end{gather} \]

The pulley and wheel start their motions with constant angular acceleration, using the equation of velocity as a function of displacement for a circular motion

\[ \bbox[#99CCFF,10px] {\omega ^{2}=\omega_{0}^{2}+2\alpha \Delta \varphi} \]

\[ \begin{gather} \omega_{A}^{2}=\omega_{0A}^{2}+2\alpha_{A}\Delta\varphi_{A}\\ \omega_{A}^{2}=\omega_{0A}^{2}+2\alpha_{A}(\varphi_{A}-\varphi_{0A})\\ \omega_{A}^{2}=0+2\times 2\times (33.5-0)\\ \omega_{A}=\sqrt{134\;}\\ \omega_{A}=11.6\;\text{rad/s} \end{gather} \]

As the pulley and wheel spin together they have the same displacement at the same speed

\[ \begin{gather} \bbox[#99CCFF,10px] {v=\omega r} \tag{II} \end{gather} \]

Applying the condition of equality to speed, we get the speed of a point P of the wheel

\[ \begin{gather} v_{A}=v_{B}\\ v_{B}=v_{A}=\omega_{A}r_{A}\\ v_{B}=\omega_{A}r_{A}\\ v_{B}=11.6\times 0.15 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {v_{B}=1.74\;\text{m/s}} \]

The magnitude of the tangential acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {a_{t}=\alpha r} \tag{III} \end{gather} \]

Applying the equality condition for acceleration, we get the acceleration of a wheel point P (Figure 2)

\[ \begin{gather} a_{tA}=a_{tB}\\ a_{tB}=a_{tA}=\alpha_{A}r_{A}\\ a_{tB}=\alpha_{A}r_{A}\\ a_{tB}=2\times 0.15\\ a_{tB}=0.30\;\text{m/s}^{2} \end{gather} \]

Figure 2

The normal acceleration (centripetal) for wheel B is given by

\[ \bbox[#99CCFF,10px] {a_{cp}=\frac{v^{2}}{r}} \]

\[ \begin{gather} a_{nB}=\frac{v_{B}^{2}}{r_{B}}\\ a_{nB}=\frac{1.74^{2}}{0.40}\\ a_{nB}=7.6\;\text{m/s}^{2} \end{gather} \]

The total acceleration will be the vector sum of the tangential and normal components above

\[ {\vec{a}}_{B}={\vec{a}}_{tB}+{\vec{a}}_{nB} \]

using the Pythagorean Theorem, the magnitude of the total acceleration will be

\[ \begin{gather} a_{B}^{2}=a_{tB}^{2}+a_{nB}^{2}\\ a_{B}^{2}=0.30^{2}+7.6^{2}\\ a_{B}^{2}=0.09+57.8\\ a_{B}=\sqrt{57.9\;} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {a_{B}\approx 7.6\;\text{m/s}^{2}} \]

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