CBSE Class 11-science Answered
Q no.53
Asked by Vivek.d8765 | 25 Jun, 2018, 11:05: AM
Expert Answer
acceleration of Centre of mass = 2.5 cm/s2
After 2 seconds, Centre of mass moves with velocity 2.5×2 = 5 cm/s
(a.1) velocity at A is twice of that of centre of mass, hence it is 10 cm/s
(a.2) velocity at B is due to the resultant velocity of linear motion i.e, 5 cm/s and
the tangential component of velocity 5cm/s due to rotational motion.
Since these two equal components are perpendicular each other resultant is 5√2 cm/s
(a.3) velocity at point of contact O is zero.
(b.1) acceleration at A is again twice of that of centre of mass, hence it is 2×2.5 = 5 cm/s2
(b.2) acceleration at B is, as explained above, 2.5×√2 = 3.54 cm/s2
(b.3) acceleration at point of contact O is zero
Answered by Thiyagarajan K | 25 Jun, 2018, 04:26: PM
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