NEET-XII-Physics
exam-1 year:2016
- #19A 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 2.0 rad ``s^{-2}``. Its net
acceleration in m ``s^{-2}`` at the end of 2.0 s is
approximately :
(1) 8.0
(2) 7.0
(3) 6.0
(4) 3.0digAnsr: 1Ans : 1
Sol. Particle at periphery will have both radial and
tangential acceleration
at = R= 0.5 × 2 = 1 m/s2
= 0 + t
= 0 + 2 × 2 = 4 rad/sec
ac = 2R = (4)2 × 0.5 = 16 × 0.5 = 8 m/s2
atotal = + = +
2 2 2 2 2
p ca a 1 8 8m/s
*In this question we have assumed the point
to be located at periphery of the disc.