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A wheel starting from rest gains an angular velocity of $10 \mathrm{rad} / \mathrm{s}$ after uniformly
accelerated for $5 \mathrm{~s}$. The total angle through which it has turned is
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accelerated for $5 \mathrm{~s}$. The total angle through which it has turned is
Solution:
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Verified Answer
The correct answer is:
$25 \mathrm{rad}$
Initial angular velocity of wheel, $\omega_{0}=0$
Final angular velocity, $\omega=10 \mathrm{rad} / \mathrm{s}$
$t=5 \mathrm{~s}$
By first equation of rotational motion,
$\begin{array}{ll}
& \omega=\omega_{0}+\alpha t \\
\Rightarrow \quad & 10=0+\alpha \times 5 \\
\Rightarrow & \alpha=\frac{10}{5}=2 \mathrm{rad} / \mathrm{s}^{2}
\end{array}$
If $\theta$ be the total angle through which wheel has turned, then from,
$\begin{aligned}
\theta &=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\
&=0 \times 5+\frac{1}{2} \times 2 \times 5^{2}=0+25 \\
\Rightarrow \quad \theta &=25 \mathrm{rad}
\end{aligned}$
Final angular velocity, $\omega=10 \mathrm{rad} / \mathrm{s}$
$t=5 \mathrm{~s}$
By first equation of rotational motion,
$\begin{array}{ll}
& \omega=\omega_{0}+\alpha t \\
\Rightarrow \quad & 10=0+\alpha \times 5 \\
\Rightarrow & \alpha=\frac{10}{5}=2 \mathrm{rad} / \mathrm{s}^{2}
\end{array}$
If $\theta$ be the total angle through which wheel has turned, then from,
$\begin{aligned}
\theta &=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\
&=0 \times 5+\frac{1}{2} \times 2 \times 5^{2}=0+25 \\
\Rightarrow \quad \theta &=25 \mathrm{rad}
\end{aligned}$
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