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:
1.The speed of the particle after $7s$ is

• A. $5m{s}^{-1}$
• B. $7m{s}^{-1}$
• C. $35m{s}^{-1}$
• D. $48m{s}^{-1}$

Answer 1

The correct option is C

$35m{s}^{-1}$

Step 1: Given data:

Tangential acceleration and radius :$a_T=5m{s}^{-2},R=0.7m$

Formula used:

Tangential acceleration: $a_T=R\alpha$ where $\alpha$ is the angular acceleration.

By equation of motion, $\omega ={\omega }_0+\alpha t$ where ${\omega }_0$is the initial angular velocity and $\omega$is the final angular velocity.

Relation between linear and angular velocity $\nu =R\omega$ where R is the radius.

Step 2: To find tangential acceleration:

$$\begin{gathered} \alpha =\frac{a_T}{R} \\ \Rightarrow \alpha =\frac{5}{0.7} \\ \Rightarrow \alpha =7.143rad{s}^{-2} \end{gathered}$$

Step 3: To find angular velocity:

We know $\omega ={\omega }_0+\alpha t$ where ${\omega }_0$is the initial angular velocity and $\omega$is the final angular velocity.

$$\begin{gathered} \mathrm{But}\mathrm{here}{\omega }_0=0 \\ \omega =\alpha t \\ \Rightarrow \omega =7.143\times 7[\because t=7s] \\ \Rightarrow \omega =50rad{s}^{-1} \end{gathered}$$

Step 4: To find the linear speed:

We know that the linear speed is given as
$$\begin{gathered} \nu =R\omega \\ \Rightarrow \nu =0.7\times 50 \\ \nu =35m{s}^{-1} \end{gathered}$$

The speed of the particle after $7s$ is $35m{s}^{-1}$.

Hence, option (c) is correct.