Question

A particle moves along a circle of radius $\left(\frac{20}{\pi }\right)m$ with tangential acceleration of constant magnitude. If the speed of the particle is $80m/s$ at the end of the Second revolution after motion has begun, the tangential acceleration is

• A. $160\pi m/s^2$
• B. $40\pi m/s^2$
• C. $40m/s^2$
• D. $640\pi m/s^2$

Answer 1

Answer: (C)

$40m/s^2$

Total angular displacement $\theta =2\times 2\pi =4\pi rad$

Final angular velocity, ${\omega }_f=v_f/r=\frac{80}{20/\pi }=4\pi rad/s$

${\omega }_f^2={\omega }_i^2+2\alpha \theta$

$(4\pi {)}^2=0^2+2\times \alpha \times 4\pi$

$\alpha =2\pi rad/s^2$

$a=\alpha r$

$=2\pi \times 20/\pi =40m/s^2$