Question

A point on the rim of a 0.750.75-m-diameter grinding wheel changes speed uniformly from 1212 m/s to 2525 m/s in 6.26.2 s. What is the angular acceleration of the grinding wheel during this interval?

Solution

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Given values:\textbf{Given values:}

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

initial linear velocity=V1=12 m/s{\text{initial linear velocity} = V_1 = 12 \text{~m/s}}

final linear velocity=V2=25 m/s{\text{final linear velocity} = V_2 = 25 \text{~m/s}}

t=6.2 s{\text{t} = 6.2 \text{~s}}

In this problem, the grinding wheel turns from linear velocity, V1V_1 to V2V_2 within a specific time, tt. Before we can solve the angular acceleration, α\alpha , of the wheel, we need to find first its linear acceleration, ata_t , using the constant acceleration equation ;

at=ΔVΔt{a_t = \dfrac{\Delta V}{\Delta t}}

at=V2V1t{a_t = \dfrac{{V_2}-{V_1}}{t}}

at=(2512) m/s6.2 s{a_t = \dfrac{(25-12) \text{~m/s}}{6.2 \text{~s}}}

at=2.1 m/s2{a_t = 2.1 ~m/s^2}

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

at=α×r{a_t = \alpha \times r}

So,

α=atr{\alpha = \dfrac{a_t}{r}}

α=2.1 m/s20.75 m2{\alpha = \dfrac{2.1 ~m/s^2}{\dfrac{0.75~m}{2}}}

α=5.6 rad/s2{\boxed{\alpha = 5.6 ~rad/s^2}}

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