A) \[160\,\,\pi \,m/{{s}^{2}}\]
B) \[40\,m/{{s}^{2}}\]
C) \[40\,\,\pi \,m/{{s}^{2}}\]
D) \[640\,\,\pi \,m/{{s}^{2}}\]
Correct Answer: B
Solution :
| Key Idea: The tangential acceleration in a circular path is the product of radius of circular path and angular acceleration. The tangential acceleration |
| \[{{a}_{T}}=r\alpha \] ...(i) |
| From 2nd equation of motion in rotational motion, |
| \[{{\omega }^{2}}=\omega _{0}^{2}+2\alpha \theta \] |
| Here, \[{{\omega }_{0}}=0,\,\omega =\frac{v}{r}=\frac{80}{20/\pi }=4\pi \,\,\,rad/s\] |
| \[\theta =2\times 2\pi \,rad\] |
| So, \[\alpha =\frac{{{\omega }^{2}}}{2\theta }=\frac{{{(4\pi )}^{2}}}{2\times (2\times 2\pi )}\] |
| \[=\frac{16{{\pi }^{2}}}{8\pi }=2\pi \] |
| Hence, from Eq. (i), we have |
| \[{{a}_{T}}=r\alpha =\frac{20}{\pi }\times 2\pi =40\,m/{{s}^{2}}\] |
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