Question

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

• A. $5{\mathrm{ms}}^{-1}$
• B. $7{\mathrm{ms}}^{-1}$
• C. $35{\mathrm{ms}}^{-1}$
• D. $48{\mathrm{ms}}^{-1}$

Answer 1

The correct option is C

$35{\mathrm{ms}}^{-1}$

Step 1: Given data:

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

Formula required:

Tangential acceleration, ${\mathrm{a}}_{\mathrm{T}}=\mathrm{R\alpha }\to \left(1\right)$ where $\mathrm{\alpha }=$ angular acceleration

We know from equations of motion , that $\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$, here $\mathrm{\omega }=$ final angular velocity and ${\mathrm{\omega }}_{0=}$ initial angular velocity.

Step 2: To calculate the angular acceleration:

The angular acceleration,

$$\begin{gathered} \mathrm{\alpha }=\frac{{\mathrm{\alpha }}_{\mathrm{T}}}{\mathrm{R}} \\ \Rightarrow \mathrm{\alpha }=\frac{5}{0.7} \\ \Rightarrow \mathrm{\alpha }=7.143{\mathrm{rads}}^{-2} \end{gathered}$$

Step 3: To calculate angular velocity:

We know, $\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$
But here ${\mathrm{\omega }}_0=0$
$$\begin{gathered} \mathrm{\omega }=\mathrm{\alpha t} \\ \Rightarrow \mathrm{\omega }=7.143*7\left[\because \mathrm{t}=7\mathrm{s}\right] \\ \Rightarrow \mathrm{\omega }=50{\mathrm{rads}}^{-1} \end{gathered}$$

Step 4: To calculate the linear speed:

We know that the linear speed is given as
$$\begin{gathered} \mathrm{v}=\mathrm{R\omega } \\ \Rightarrow \mathrm{v}=0.7*50 \\ \Rightarrow \mathrm{v}=35{\mathrm{ms}}^{-1} \end{gathered}$$

The speed of the particle after $7$sec is $35{\mathrm{ms}}^{-1}$.

Hence, option (c) is correct.