NEET Solved Paper 2016 Question 19
Question: A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $ 2\text{.0 rad }{s^{-2}}. $ Its net acceleration in $ m{s^{-2}} $ at the end of 2.0 s is approximately :
Options:
A) 8.0
B) 7.0
C) 6.0
D) 3.0
Show Answer
Answer:
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=4rad/\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.
