A wheel is subjected to uniform angular accel | vedclass

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Question

(c) Using relation $	heta = {\omega _0}t + \frac{1}{2}a{t^2}$

${	heta _1} = \frac{1}{2}(\alpha ){(2)^2} = 2\alpha $&hellip;$(i)$ (As ${\omega _0}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c) Using relation $	heta = {\omega _0}t + \frac{1}{2}a{t^2}$

${	heta _1} = \frac{1}{2}(\alpha ){(2)^2} = 2\alpha $&hellip;$(i)$ (As ${\omega _0} = 0,t = 2\,\sec $)

Now using same equation for $t = 4 \,sec$, $\omega_0 = 0$

${	heta _1} + {	heta _2} = \frac{1}{2}\alpha {(4)^2} = 8\alpha $&hellip;$(ii)$

From $(i)$ and $(ii)$,

${	heta _1} = 2\alpha $and${	heta _2} = 6\alpha $    $\frac{{{	heta _2}}}{{{	heta _1}}} = 3$

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first $2$ sec, it rotates through an angle ${\theta _1}$. In the next $2$ sec, it rotates through an additional angle ${\theta _2}$. The ratio of ${\theta _2}\over{\theta _1}$ is

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