A) 8.0
B) 7.0
C) 6.0
D) 3.0
Correct Answer: A
Solution :
Particle at periphery will have both radial and tangential acceleration \[{{a}_{t}}=R\alpha =0.5\times 2=1\,m/{{s}^{2}}\] \[\omega ={{\omega }_{0}}+\alpha t\] \[\omega =0+2\times 2=4\,\,rad/\sec \] \[{{a}_{c}}={{\omega }^{2}}R={{(4)}^{2}}\times 0.5=16\times 0.5=8\,m/{{s}^{2}}\] \[{{a}_{total}}=\sqrt{a_{p}^{2}+a_{c}^{2}}=\sqrt{{{1}^{2}}+{{8}^{2}}}\approx \,\,8m/{{s}^{2}}\] *In this question we have assumed the point to be located at periphery of the disc.You need to login to perform this action.
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