Answer to Question #172816 in Mechanical Engineering for clar

Question

A 45.3 - cm radius wheel is uniformly speeding up from 146 rev/min to 621 rev/min within 26.5 s. Find the tangential acceleration.

Answer 1

Here revolution per min is given

$N_1=$ 146 rev/min , $\omega_1= \frac{2\pi N_1}{60}= \frac{2\times 3.14\times 146}{60}=15.28 \frac{rad}{s}$

$N_2= \omega_2= \frac{2\pi N_2}{60}= \frac{2\times 3.14\times 621}{60}=64.998 \frac{rad}{s}$

t=26.5 s,

using first equation of rotation motion

$\omega_2=\omega_1+ \alpha t$

$\alpha=1.876 \frac{rad}{s^2}$

$\alpha$ = angular acceleration

tangential acceleration,

$a= r \alpha=0.85 \frac{m}{s^2}$