On the application of a constant torque, a wheel is turned from rest through $400$ radians in $10$ s. Find angular acceleration.

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Solution

The initial angular velocity $ω_{0} = 0 s e c$ and angular velocity after 10 seconds sec $ω = \frac{400}{10} = 40 \frac{r a d}{s e c}$
Therefore angular acceleration sec $α = \frac{ω - ω_{0}}{t} = \frac{40 - 0}{10} = 4 \frac{r a d}{{s e c}^{2}}$

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On the application of a constant torque, a wheel is turned from rest through 400 radians in 10s. Find angular acceleration.

The initial angular velocity ω0=0sec and angular velocity after 10 seconds sec ω=40010=40radsecTherefore angular acceleration sec α=ω−ω0t=40−010=4radsec2

On the application of a constant torque, a wheel is turned from rest through $400$ radians in $10$ s. Find angular acceleration.

The initial angular velocity $ω_{0} = 0 s e c$ and angular velocity after 10 seconds sec $ω = \frac{400}{10} = 40 \frac{r a d}{s e c}$
Therefore angular acceleration sec $α = \frac{ω - ω_{0}}{t} = \frac{40 - 0}{10} = 4 \frac{r a d}{{s e c}^{2}}$