A particle moves along a circle of radius R . Its speed v varies with distance x covered along the circle `v=ksqrt(x)` ,where is a constant .if R =1.0 m and x= `0.5` m , find the angle between the net acceleration vector and the tangential acceleration vector .

Options

A. `45^(@)`
B. `60^(@)`
C. `90^(@)`
D. `150^(@)`

Answer

A

Explanation

Tangential accelerations is `a_(t) =(dv)/(dt) =(dv)/(dt) (dx)/(dx)=("v dv")/(dx)`
`rArr a_(t) dx = "v dv"`
`a_(t) int_(0)^(H) dx = int_(0)^(v) " v dv"`
`rArr a_(t)x =(v^(2))/2`
`rArr a_(t) =(v^(2))/(2x) =((ksqrt(x))^(2))/(2x)=(k^(2))/2`
Centripetal acceleration is
`a_(c ) =(v^(2))/R = (k^(2)x)/R`
Angle `theta ` between net acceleration and tangential acceleration is given by
`tan theta =(a_(c ))/(a_(t)) =(k^(2)x//r)/(k^(2)//2)=x/(2R)`
Putting x = `0.5` m and R = 1.0 m .We get
`tan theta = (2 xx 0.5)/(1.0) = 1 rArr theta = 45^(@)`