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 \[{{a}_{T}}=r\alpha =\frac{20}{\pi }\times 2\pi =40\,m/{{s}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
