A particle of mass 'm' moves along a circle of radius 'r' with constant tangential acceleration. If kinetic…

A particle of mass 'm' moves along a circle of radius 'r' with constant tangential acceleration. If kinetic energy 'E' of the particle becomes three times by the end of third revolution after beginning of the motion then the magnitude of tangential acceleration is
  1. $\frac{\mathrm{E}}{12 \pi \mathrm{rm}}$
  2. $\frac{\mathrm{E}}{3 \pi \mathrm{rm}}$
  3. $\frac{\mathrm{E}}{6 \pi \mathrm{rm}}$
  4. $\frac{\mathrm{E}}{24 \pi \mathrm{rm}}$

Solution

$\mathrm{E}_{1}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mr}^{2} \omega_{1}^{2}$ $E_{2}=\frac{3}{2} m r^{2} \omega_{1}^{2}$ $\frac{E_{2}}{E_{1}}=3 \quad \therefore E_{2}=3 E_{1}$ $\frac{1}{2} \mathrm{mr}^{2} \omega^{2}=3 \frac{1}{2} \mathrm{~m} \omega_{0}^{2} \mathrm{r}^{2}$ $\omega^{2}=3 \omega_{0}^{2}$ $\frac{1}{2} m \omega_{0}^{2} r^{2}=E$ $\omega_{0}^{2}=\frac{2 \mathrm{E}}{\mathrm{mr}^{2}}$ $\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$ $3 \omega_{0}^{2}=\omega_{0}^{2}+2 \alpha(3.2 \pi)$ $2 \omega_{0}^{2}=12 \alpha \pi$ $\alpha=\frac{\omega_{0}^{2}}{6 \pi}=\frac{2 \mathrm{E}}{\mathrm{mr}^{2}} \times \frac{1}{6 \pi}=\frac{\mathrm{E}}{3 \pi \mathrm{mr}^{2}}$ But $\mathrm{a}=r \alpha=r \times \frac{\mathrm{E}}{3 \pi \mathrm{mr}^{2}}=\frac{\mathrm{E}}{3 \pi \mathrm{mr}}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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