
A particle of mass m is moving in a circular path of constant radius r such that centripetal acceleration is varying with time t as \[{{k}^{2}}r{{t}^{2}}\], where k is a constant. The power delivered to the particle by the force acting on it is:
A. \[{{m}^{2}}{{k}^{2}}{{r}^{2}}{{t}^{2}}\]
B. \[m{{k}^{2}}{{r}^{2}}t\]
C. \[m{{k}^{2}}r{{t}^{2}}\]
D. \[mk{{r}^{2}}t\]
Answer
446.2k+ views
Hint: The centripetal acceleration of the body is directed towards the center along the radius of the circular path. The tangential acceleration of the body is directed along the tangent to the circular path.
Complete step by step solution:
If a body is moving in a circular path of radius $R$ with linear speed $v$
Then the centripetal acceleration is given as,
${{a}_{c}}=\dfrac{{{v}^{2}}}{R}\ldots \ldots \left( i \right)$
It is given that centripetal acceleration, ${{a}_{c}}={{k}^{2}}r{{t}^{2}}\ldots \ldots \left( ii \right)$
From equation $\left( i \right)$ and $\left( ii \right)$
$\begin{align}
& \dfrac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}} \\
& {{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}} \\
& v=krt\ldots \ldots \left( iii \right)
\end{align}$
Tangential acceleration given as,
${{a}_{t}}=\dfrac{dv}{dt}$
From equation $\left( iii \right)$,
$\begin{align}
& {{a}_{t}}=\dfrac{d}{dt}\left( krt \right) \\
& =kr\ldots \ldots \left( iv \right)
\end{align}$
Tangential force acting on the particle,
$\begin{align}
& F=m{{a}_{t}} \\
& =mkr
\end{align}$
Instantaneous power delivered by the body is given as,
$\begin{align}
& P=\overrightarrow{F}\centerdot \overrightarrow{v} \\
& =Fv\left( \cos \theta \right)
\end{align}$
The angle between the linear velocity and the tangential force is $\theta =0{}^\circ $
Power delivered,
$\begin{align}
& P=Fv\left( \cos 0{}^\circ \right) \\
& =Fv \\
& =\left( mkr \right)\left( krt \right) \\
& =m{{k}^{2}}{{r}^{2}}t
\end{align}$
Therefore, the power delivered by is $m{{k}^{2}}{{r}^{2}}t$.
Note: Power delivered by the force is the scalar product of the force and the velocity. Power delivered during uniform circular motion is zero.
Complete step by step solution:
If a body is moving in a circular path of radius $R$ with linear speed $v$
Then the centripetal acceleration is given as,
${{a}_{c}}=\dfrac{{{v}^{2}}}{R}\ldots \ldots \left( i \right)$
It is given that centripetal acceleration, ${{a}_{c}}={{k}^{2}}r{{t}^{2}}\ldots \ldots \left( ii \right)$
From equation $\left( i \right)$ and $\left( ii \right)$
$\begin{align}
& \dfrac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}} \\
& {{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}} \\
& v=krt\ldots \ldots \left( iii \right)
\end{align}$
Tangential acceleration given as,
${{a}_{t}}=\dfrac{dv}{dt}$
From equation $\left( iii \right)$,
$\begin{align}
& {{a}_{t}}=\dfrac{d}{dt}\left( krt \right) \\
& =kr\ldots \ldots \left( iv \right)
\end{align}$
Tangential force acting on the particle,
$\begin{align}
& F=m{{a}_{t}} \\
& =mkr
\end{align}$
Instantaneous power delivered by the body is given as,
$\begin{align}
& P=\overrightarrow{F}\centerdot \overrightarrow{v} \\
& =Fv\left( \cos \theta \right)
\end{align}$
The angle between the linear velocity and the tangential force is $\theta =0{}^\circ $
Power delivered,
$\begin{align}
& P=Fv\left( \cos 0{}^\circ \right) \\
& =Fv \\
& =\left( mkr \right)\left( krt \right) \\
& =m{{k}^{2}}{{r}^{2}}t
\end{align}$
Therefore, the power delivered by is $m{{k}^{2}}{{r}^{2}}t$.
Note: Power delivered by the force is the scalar product of the force and the velocity. Power delivered during uniform circular motion is zero.
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