MHT CET · Physics · Work Power Energy
A particle at rest starts moving with a constant angular acceleration of \(4 \mathrm{rad} / \mathrm{s}^2\) in a circular path. The time at which magnitudes of its centripetal acceleration and tangential acceleration will be equal, is (in second)
- A \(\frac{1}{4}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given that, \(\alpha=4 \mathrm{rad} / \mathrm{s}^2\)
Centripetal (radial) acceleration, \(\mathrm{a}_{\mathrm{r}}=\mathrm{r} \omega^2\)
Tangential acceleration, \(a_t=r \alpha\)
If \(a_r=a_t\), then \(r \omega^2=r \alpha\)
\(\begin{array}{ll} \therefore & \omega^2=\alpha=4 \\ \therefore & \omega=\sqrt{4}=2 \mathrm{rad} / \mathrm{s} \\ \therefore & \text { But, } \omega=\omega_0+\alpha \mathrm{t}=0+\alpha \mathrm{t}=\alpha \mathrm{t} \\ \therefore & 2=4 \mathrm{t} \\ \therefore & \mathrm{t}=\frac{1}{2} \mathrm{~s} \end{array}\)
Centripetal (radial) acceleration, \(\mathrm{a}_{\mathrm{r}}=\mathrm{r} \omega^2\)
Tangential acceleration, \(a_t=r \alpha\)
If \(a_r=a_t\), then \(r \omega^2=r \alpha\)
\(\begin{array}{ll} \therefore & \omega^2=\alpha=4 \\ \therefore & \omega=\sqrt{4}=2 \mathrm{rad} / \mathrm{s} \\ \therefore & \text { But, } \omega=\omega_0+\alpha \mathrm{t}=0+\alpha \mathrm{t}=\alpha \mathrm{t} \\ \therefore & 2=4 \mathrm{t} \\ \therefore & \mathrm{t}=\frac{1}{2} \mathrm{~s} \end{array}\)
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