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A stone tied to the end of a string of $1 \mathrm{~m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what would be the magnitude and direction of acceleration of the stone?
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Verified Answer
The correct answer is:
$\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre.
Acceleration will be $\pi 2 \mathrm{~m} / \mathrm{s} 2$ radially inwards. Radial acceleration or centripetal accceleration is given by:
$\mathrm{ac}=\mathrm{v} 2 \mathrm{r}=\omega 2 \mathrm{r}$
Frequency of revolution is,
$\begin{aligned}
& \mathrm{f}=2244=12 \mathrm{~s}-1 \\
& \Rightarrow \omega=2 \pi \mathrm{f}=\pi \mathrm{rad} / \mathrm{s} \\
& \therefore \mathrm{ac}=\omega 2 \mathrm{r}=\pi 2 \times 1=\pi 2 \mathrm{~m} / \mathrm{s} 2
\end{aligned}$
Tangential acceleration is the rate of change of speed w.r.t time:
$\Rightarrow a_t=\frac{d v}{d t}=0$
$\because$ speed is constant
Hence net acceleration of stone $\vec{a}$ given by:
$\begin{aligned}
& \vec{a}=\vec{a}_c+\vec{a}_t=\vec{a}_c=\pi^2 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore|\vec{a}|=\left|\vec{a}_c\right|=\pi^2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
Direction is along the radius towards the centre i.e. radially inwards.
$\mathrm{ac}=\mathrm{v} 2 \mathrm{r}=\omega 2 \mathrm{r}$
Frequency of revolution is,
$\begin{aligned}
& \mathrm{f}=2244=12 \mathrm{~s}-1 \\
& \Rightarrow \omega=2 \pi \mathrm{f}=\pi \mathrm{rad} / \mathrm{s} \\
& \therefore \mathrm{ac}=\omega 2 \mathrm{r}=\pi 2 \times 1=\pi 2 \mathrm{~m} / \mathrm{s} 2
\end{aligned}$
Tangential acceleration is the rate of change of speed w.r.t time:
$\Rightarrow a_t=\frac{d v}{d t}=0$
$\because$ speed is constant
Hence net acceleration of stone $\vec{a}$ given by:
$\begin{aligned}
& \vec{a}=\vec{a}_c+\vec{a}_t=\vec{a}_c=\pi^2 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore|\vec{a}|=\left|\vec{a}_c\right|=\pi^2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
Direction is along the radius towards the centre i.e. radially inwards.
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