Question

The speed of a particle traveling in a circle of radius $20cm$ increases uniformly from $6.0m/s$ to $8.0m/s$ in $4.0s$, find the angular acceleration.

• A. $2rad/s^2$
• B. $2.5rad/s^2$
• C. $5rad/s^2$
• D. $1.5rad/s^2$

Answer 1

Answer: (B)

$2.5rad/s^2$

Answer is B.
Since speed increases uniformly, average tangential acceleration is equal to instantaneous tangential acceleration.
Therefore, The instantaneous tangential acceleration is given by $a_t=\frac{dv}{dt}=\frac{v_2-v_1}{t_2-t_1}$
$=\frac{8.0-6.0}{4.0}m/s^2=0.5m/s^2$
The angular acceleration is $\alpha =a_1/r$
$=\frac{0.5m/s^2}{20cm}=2.5rad/s^2$.
Hence, the angular acceleration is $2.5rad/s^2$.