Question

A particle is moving along a circular path of radius $0.7\mathrm{m}$ with tangential acceleration of $5{\mathrm{ms}}^{-2}$ The particle is initially at rest.
Based on above information, answer the following question:

4.The number of revolution made by the particle in $7\text{s}$ is

• A. $27.8\mathrm{rev}$
• B. $164.8\mathrm{rev}$
• C. $165.52\mathrm{rev}$
• D. $96.85\mathrm{rev}$

Answer 1

The correct option is A

$27.8\mathrm{rev}$

Step 1: Given data:

Tangential acceleration and radius: ${\mathrm{\alpha }}_{\mathrm{T}}=5{\mathrm{ms}}^{-2},\mathrm{R}=0.7\mathrm{m}$

Formula used:

Tangential acceleration, ${\mathrm{\alpha }}_{\mathrm{T}}=\mathrm{R\alpha }$ where $\mathrm{\alpha }=$ angular acceleration.

The angular displacement is given as $\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ where ${\mathrm{\omega }}_0=$ initial angular velocity and here ${\mathrm{\omega }}_0=0$

Step 2: To find angular acceleration:
$$\begin{gathered} \therefore \mathrm{\alpha }=\frac{{\mathrm{\alpha }}_{\mathrm{T}}}{\mathrm{R}} \\ \Rightarrow \mathrm{\alpha }=\frac{5}{0.7} \\ \Rightarrow \mathrm{\alpha }=7.143{\mathrm{rads}}^{-2} \end{gathered}$$

Step 3: To find the angular displacement:
We know that the angular displacement is given as
$\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$
Where ${\mathrm{\omega }}_0=$ initial angular velocity and here ${\mathrm{\omega }}_0=0$
$$\begin{gathered} \therefore \mathrm{\theta }=\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow \mathrm{\theta }=\frac{1}{2}*7.143*{\left(7\right)}^2\mathrm{rad} \\ \Rightarrow \mathrm{\theta }=175\mathrm{rad} \end{gathered}$$

Step 4: To find number of revolution:

We know that the number of revolution in $7\mathrm{s}$ is given as
$$\begin{gathered} =\frac{\mathrm{\theta }}{2\mathrm{\pi }} \\ =\frac{175}{2\mathrm{\pi }} \\ \approx 27.8\mathrm{rev} \end{gathered}$$

The number of revolution made by the particle in $7\text{s}$ is $27.8\mathrm{rev}$

Hence, option (a) is correct.