Question

# A point on the rim of a $0.75$-m-diameter grinding wheel changes speed uniformly from $12$ m/s to $25$ m/s in $6.2$ s. What is the angular acceleration of the grinding wheel during this interval?

Solution

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$\textbf{Given values:}$

${\text{diameter} = d_w = 0.75 \text{~m}}$

${\text{initial linear velocity} = V_1 = 12 \text{~m/s}}$

${\text{final linear velocity} = V_2 = 25 \text{~m/s}}$

${\text{t} = 6.2 \text{~s}}$

In this problem, the grinding wheel turns from linear velocity, $V_1$ to $V_2$ within a specific time, $t$. Before we can solve the angular acceleration, $\alpha$ , of the wheel, we need to find first its linear acceleration, $a_t$ , using the constant acceleration equation ;

${a_t = \dfrac{\Delta V}{\Delta t}}$

${a_t = \dfrac{{V_2}-{V_1}}{t}}$

${a_t = \dfrac{(25-12) \text{~m/s}}{6.2 \text{~s}}}$

${a_t = 2.1 ~m/s^2}$

Now solving for the angular acceleration, $\alpha$ , using the angular acceleration equation ;

${a_t = \alpha \times r}$

So,

${\alpha = \dfrac{a_t}{r}}$

${\alpha = \dfrac{2.1 ~m/s^2}{\dfrac{0.75~m}{2}}}$

${\boxed{\alpha = 5.6 ~rad/s^2}}$

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