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A wheel starts rotating from rest at time $t=0$ with a angular acceleration of 50 radians $/ \mathrm{s}^2$. The angular acceleration $(\alpha)$ decreases to zero value after 5 seconds. During this interval, $\alpha$ varies according to the equation
$\alpha=\alpha_0\left(1-\frac{t}{5}\right)$
The angular velocity at $t=5 \mathrm{~s}$ will be
Options:
$\alpha=\alpha_0\left(1-\frac{t}{5}\right)$
The angular velocity at $t=5 \mathrm{~s}$ will be
Solution:
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Verified Answer
The correct answer is:
$125 \mathrm{rad} / \mathrm{s}$
$\alpha=\alpha_0\left(1-\frac{t}{5}\right)$
At $t=0, \alpha=\alpha_0 \quad \therefore \alpha_0=50 \mathrm{rad} / \mathrm{s}^2$
$\frac{d \omega}{d t}=\alpha_0\left(1-\frac{t}{5}\right)$
$\begin{aligned} & \therefore \int_0^\omega d \omega=\alpha_0 \int_0^5\left(1-\frac{t}{5}\right) d t \Rightarrow \omega=\alpha_0\left[t-\frac{t^2}{10}\right]_0^5 \\ & =50\left(5-\frac{25}{10}\right) \mathrm{rad} / \mathrm{s}=125 \mathrm{rad} / \mathrm{s}\end{aligned}$
At $t=0, \alpha=\alpha_0 \quad \therefore \alpha_0=50 \mathrm{rad} / \mathrm{s}^2$
$\frac{d \omega}{d t}=\alpha_0\left(1-\frac{t}{5}\right)$
$\begin{aligned} & \therefore \int_0^\omega d \omega=\alpha_0 \int_0^5\left(1-\frac{t}{5}\right) d t \Rightarrow \omega=\alpha_0\left[t-\frac{t^2}{10}\right]_0^5 \\ & =50\left(5-\frac{25}{10}\right) \mathrm{rad} / \mathrm{s}=125 \mathrm{rad} / \mathrm{s}\end{aligned}$
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