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  4. A grindstone increases in angular speed from 4.00 rad / s to 12.00 rad / s in 4.00 s. Through what angle does it turn during that time interval if the angular acceleration is constant?

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Q. A grindstone increases in angular speed from $4.00 \,rad / s$ to $12.00 \,rad / s$ in $4.00 s$. Through what angle does it turn during that time interval if the angular acceleration is constant?

System of Particles and Rotational Motion

Solution:

The angular displacement will be
$\Delta \theta =\omega_{ avg } \Delta t=\left(\frac{\omega_{f}+\omega_{i}}{2}\right) \Delta t $
$ =\left(\frac{12.00\, rad / s +4.00 \,rad / s }{2}\right)(4.00 \,s )=32.0 \,rad $