# A particle starts from rest and its angular displacement (in radians) is given by $\theta =\dfrac{{{t}^{2}}}{20}+\dfrac{t}{5}$. If the angular velocity at the end of $t=4s$ is $'k'$. Then, what will be the value of $'5k'$ ?

Last updated date: 20th Mar 2023

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Answer

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Hint: The angular velocity as a function of time can be found out by differentiating angular displacement $\theta $ with respect to time.

Before proceeding with the question, we must know that the angular velocity (which is generally denoted by $\omega $) as a function of time can be found by simply differentiating the angular displacement $\theta $ with respect to time. Mathematically, we get,

$\omega =\dfrac{d\theta }{dt}............\left( 1 \right)$

Here, in this formula, $\theta $ should be a function of time $t$.

In this question, it is given that $\theta =\dfrac{{{t}^{2}}}{20}+\dfrac{t}{5}$. Substituting $\theta =\dfrac{{{t}^{2}}}{20}+\dfrac{t}{5}$ in equation $\left( 1 \right)$, we can find angular velocity as a function of time.

$\begin{align}

& \omega =\dfrac{d\left( \dfrac{{{t}^{2}}}{20}+\dfrac{t}{5} \right)}{dt} \\

& \Rightarrow \omega =\dfrac{d\left( \dfrac{{{t}^{2}}}{20} \right)}{dt}+\dfrac{d\left( \dfrac{t}{5} \right)}{dt} \\

& \Rightarrow \omega =\dfrac{1}{20}\dfrac{d\left( {{t}^{2}} \right)}{dt}+\dfrac{1}{5}\dfrac{d\left( t \right)}{dt}...............\left( 2 \right) \\

\end{align}$

In differentiation, we have a formula,

$\dfrac{d\left( {{t}^{2}} \right)}{dt}=2t...........\left( 3 \right)$

Substituting $\dfrac{d\left( {{t}^{2}} \right)}{dt}=2t$ from equation$\left( 3 \right)$ in equation $\left( 2 \right)$, we get,

$\begin{align}

& \omega =\dfrac{1}{20}\left( 2t \right)+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{t}{10}+\dfrac{1}{5}...........\left( 4 \right) \\

\end{align}$

In the question, it is given that the angular velocity $\omega $ at $t=4s$ is equal to $k$. Substituting $t=4s$in equation $\left( 4 \right)$, we get,

$\begin{align}

& \omega =\dfrac{4}{10}+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{2}{5}+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{3}{5} \\

\end{align}$

This angular velocity which we got in the above equation is equal to $k$, so, we can say,

$k=\dfrac{3}{5}..........\left( 5 \right)$

In the question, we are asked to find out the value of $5k$. So, using equation $\left( 5 \right)$, the value of $5k$ is equal to,

$\begin{align}

& 5k=5\left( \dfrac{3}{5} \right) \\

& \Rightarrow 5k=3 \\

\end{align}$

Hence, the answer is $3$.

Note: There is a possibility that one may commit a mistake while finding the angular velocity $\omega $ as a function of time. Sometimes, we integrate the angular displacement $\theta $ with respect to time to find the angular velocity instead of differentiating the angular displacement. So one must remember that angular velocity is found by differentiating the angular displacement function with respect to time

Before proceeding with the question, we must know that the angular velocity (which is generally denoted by $\omega $) as a function of time can be found by simply differentiating the angular displacement $\theta $ with respect to time. Mathematically, we get,

$\omega =\dfrac{d\theta }{dt}............\left( 1 \right)$

Here, in this formula, $\theta $ should be a function of time $t$.

In this question, it is given that $\theta =\dfrac{{{t}^{2}}}{20}+\dfrac{t}{5}$. Substituting $\theta =\dfrac{{{t}^{2}}}{20}+\dfrac{t}{5}$ in equation $\left( 1 \right)$, we can find angular velocity as a function of time.

$\begin{align}

& \omega =\dfrac{d\left( \dfrac{{{t}^{2}}}{20}+\dfrac{t}{5} \right)}{dt} \\

& \Rightarrow \omega =\dfrac{d\left( \dfrac{{{t}^{2}}}{20} \right)}{dt}+\dfrac{d\left( \dfrac{t}{5} \right)}{dt} \\

& \Rightarrow \omega =\dfrac{1}{20}\dfrac{d\left( {{t}^{2}} \right)}{dt}+\dfrac{1}{5}\dfrac{d\left( t \right)}{dt}...............\left( 2 \right) \\

\end{align}$

In differentiation, we have a formula,

$\dfrac{d\left( {{t}^{2}} \right)}{dt}=2t...........\left( 3 \right)$

Substituting $\dfrac{d\left( {{t}^{2}} \right)}{dt}=2t$ from equation$\left( 3 \right)$ in equation $\left( 2 \right)$, we get,

$\begin{align}

& \omega =\dfrac{1}{20}\left( 2t \right)+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{t}{10}+\dfrac{1}{5}...........\left( 4 \right) \\

\end{align}$

In the question, it is given that the angular velocity $\omega $ at $t=4s$ is equal to $k$. Substituting $t=4s$in equation $\left( 4 \right)$, we get,

$\begin{align}

& \omega =\dfrac{4}{10}+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{2}{5}+\dfrac{1}{5} \\

& \Rightarrow \omega =\dfrac{3}{5} \\

\end{align}$

This angular velocity which we got in the above equation is equal to $k$, so, we can say,

$k=\dfrac{3}{5}..........\left( 5 \right)$

In the question, we are asked to find out the value of $5k$. So, using equation $\left( 5 \right)$, the value of $5k$ is equal to,

$\begin{align}

& 5k=5\left( \dfrac{3}{5} \right) \\

& \Rightarrow 5k=3 \\

\end{align}$

Hence, the answer is $3$.

Note: There is a possibility that one may commit a mistake while finding the angular velocity $\omega $ as a function of time. Sometimes, we integrate the angular displacement $\theta $ with respect to time to find the angular velocity instead of differentiating the angular displacement. So one must remember that angular velocity is found by differentiating the angular displacement function with respect to time

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