A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure
(##HCV_VOL2_C38_E01_033_Q01##)

Options

A. Minimum value of `v_(0)` for the rod to rotate by an angle `(pi)/(2) is 2sqrt(gL)` .
B. Angular acceleration of the rod when the rod is horizontal is `(9g)/(8L)` .
C. Force applied by the axis on the rod in the horizontal state is `5 mg//6` .
D. for a small velue of `v_(0)` the rod performes small oscillations with a period of `(4)/(3) pi sqrt((l)/(g))`

Answer

A, B, C

Explanation

A thin ………….
(A) Conserving angular momentum about A
`mv_(0)L=((mL^(2))/(3)+mL^(2)) omega`
`implies omega = (3v_(0))/(4L)`
Now, applying energy conservation
`(1)/(2).(4mL^(2))(3)omega^(2)=mgL+(mgL)/(2)`
Put `omega`
`:. v_(0)= 2sqrt(gL)`

`mgL+(mgL)/(2)=(4mL^(2))/(3).alpha`
`alpha=(9g)/(8L)`

`mg+mg-N=2ma_(COM)`
`N=5mg//16`
(D) `T = 2pisqrt((I)/(mgL_("Distance of COM from axis")))`
`(2)/(3)pi sqrt((2l)/(g))`