pls solve

Asked by sulaikhasulu393 | 29th May, 2020, 10:00: PM

Expert Answer:

Linear velocity of the sphere is Vo and angular velocity about center is Vo/2R. We have to calculate the final translational velocity of the sphere when nit starts pure rolling.
A n g u l a r space v e l o c i t y equals fraction numerator V subscript 0 over denominator 2 R end fraction
T r a n s l a t i o n a l space v e l o c i t y equals V subscript 0
A s space V subscript 0 greater than fraction numerator V subscript 0 over denominator 2 R end fraction comma space t h e r e space w i l l space b e space s l i p p i n g space i n i t i a l l y space a f t e r space t h a t space t h e space p u r e space r o l l i n g space s t a r t
a subscript c o m end subscript equals f over m
V subscript f i n a l end subscript equals u minus a subscript c t
V subscript f i n a l end subscript equals v subscript 0 minus f over M t minus negative negative negative negative negative negative negative negative left parenthesis 1 right parenthesis
A s space w e space k n o w
capital gamma equals I alpha equals f R
alpha equals fraction numerator f R over denominator begin display style 2 over 5 end style M R squared end fraction
A n g u l a r space v e l o c i t y space a t t a i n e d space a f t e r space t i m e space t
omega equals omega subscript 0 plus alpha t
omega equals omega subscript 0 plus fraction numerator f R over denominator begin display style 2 over 5 M R squared end style end fraction t
omega equals V subscript 0 over 2 plus fraction numerator f R over denominator begin display style 2 over 5 M R squared end style end fraction t
omega equals fraction numerator V subscript 0 over denominator 2 R end fraction plus fraction numerator 5 f over denominator begin display style 2 M R end style end fraction t
omega R equals V subscript 0 over 2 plus fraction numerator 5 f t over denominator begin display style 2 M end style end fraction minus negative negative negative negative left parenthesis 2 right parenthesis
U sin g space e q u a t i o n space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis
fraction numerator 5 f over denominator 2 M end fraction t equals 5 over 2 v subscript 0 minus 5 over 2 V subscript f i n a l end subscript equals omega R minus V subscript 0 over 2
5 over 2 v subscript 0 minus 5 over 2 V subscript f i n a l end subscript equals V subscript f i n a l end subscript minus V subscript 0 over 2
3 v subscript 0 equals 7 over 2 V subscript f i n a l end subscript
V subscript f i n a l end subscript equals fraction numerator 6 v subscript 0 over denominator 7 end fraction

Answered by Utkarsh Lokhande | 30th May, 2020, 11:48: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.