KCET · Physics · Motion In Two Dimensions
A particle is in uniform circular motion. The equation of its trajectory is given by \((x-2)^2+y^2=25\), where x and y are in meter. The speed of the particle is \(2 \mathrm{~ms}^{-1}\), when the particle attains the lowest ' \(y\) ' co-ordinate, the acceleration of the particle is (in \(\mathrm{ms}^{-2}\) )
- A \(0.4 \hat{\mathrm{j}}\)
- B \(0.8 \hat{\mathrm{i}}\)
- C \(0.8 \hat{\mathrm{j}}\)
- D \(0.4 \hat{\mathrm{i}}\)
Answer & Solution
Correct Answer
(C) \(0.8 \hat{\mathrm{j}}\)
Step-by-step Solution
Detailed explanation
\((x-2)^2+y^2=25\)
Equation of circle center \(\Rightarrow(2,0)\)
\(\begin{aligned}
& r^2=25 \\
& r=5 \\
& a_c=\frac{v^2}{R}=\frac{2^2}{5} \\
& =0.8 \mathrm{~m} / \mathrm{s}^2 \\
& \overrightarrow{\mathrm{a}_{\mathrm{c}}}=0.8 \hat{\mathrm{i}} \mathrm{~m} / \mathrm{s}^2
\end{aligned}\)
Equation of circle center \(\Rightarrow(2,0)\)
\(\begin{aligned}
& r^2=25 \\
& r=5 \\
& a_c=\frac{v^2}{R}=\frac{2^2}{5} \\
& =0.8 \mathrm{~m} / \mathrm{s}^2 \\
& \overrightarrow{\mathrm{a}_{\mathrm{c}}}=0.8 \hat{\mathrm{i}} \mathrm{~m} / \mathrm{s}^2
\end{aligned}\)
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