Question

A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels

Answer 1

Answer on Question #63187, Physics / Solid State Physics

A powerful motorcycle can accelerate from 0 to $30.0\mathrm{m / s}$ (about $108~\mathrm{km / h}$) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels?

Solution:

The linear acceleration is

$$a _ {t} = \Delta v / \Delta t = 30.0 \, (\mathrm{m/s}) / 4.20 \, (\mathrm{s}) = 7.14 \, \mathrm{m/s^2}.$$

We also know the radius of the wheels.

Entering the values for $a_{t}$ and $r$ into $\alpha = a_{t} / r$, we get

$$\alpha = a _ {t} / r = 7.14 \, (\mathrm{m/s^2}) / 0.320 \, (\mathrm{m}) = 22.3 \, \mathrm{rad/s^2}$$

Answer: 22.3 rad/s²

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