Question

A particle is revolving in a circle. Its angular speed increases from 2 rad/s to $40$ rad/s in $19$ sec. The radius of the circle is $20cm$. Compute the ratio of its centripetal acceleration to its tangential acceleration at the end of 40s.

• A. $8:1$
  
• B. $1:1$
  
• C. $800:1$
  
• D. $700:1$
  

Answer 1

The correct option is C $800:1$

Given : $r=0.2m$, $w_o=2rad/s$, $w_t=40rad/s$ and $t=19s$

Tangential acceleration $a_t$ $=r\alpha =r\left(\frac{{\omega }_t-{\omega }_0}{t}\right)$

$a_t=0.20\times \frac{40-2}{19}=0.4rad/s^2$

Centripetal acceleration

${\Rightarrow a}_c={\omega }^{2}r={\left(40\right)}^2\left(0.20\right)=320m/s^2$

Ultimately ratio $\frac{a_c}{a_t}=\frac{320}{0.4}=800$